The method laid out below is the one described in Section 6.3.3 of Davidson and Hinckley (1997), Bootstrap Methods and Their Application. Thanks to Glen_b and his comment here. Given that there were several questions on Cross Validated on this topic, I thought it was worth writing up.

The linear regression model is: \begin{align} Y_i &= X_i\beta+\epsilon_i \end{align}

We have data, $i=1,2,\ldots,N$, which we use to estimate the $\beta$ as: \begin{align} \hat{\beta}_{\text{OLS}} &= \left( X'X \right)^{-1}X'Y \end{align}

Now, we want to predict what $Y$ will be for a new data point, given that we know $X$ for it. This is the prediction problem. Let's call the new $X$ (which we know) $X_{N+1}$ and the new $Y$ (which we would like to predict), $Y_{N+1}$. The usual prediction (if we are assuming that the $\epsilon_i$ are iid and uncorrelated with $X$) is: \begin{align} Y^p_{N+1} &= X_{N+1}\hat{\beta}_{\text{OLS}} \end{align}

The forecast error made by this prediction is: \begin{align} e^p_{N+1} &= Y_{N+1}-Y^p_{N+1} \end{align}

We can re-write this equation like: \begin{align} Y_{N+1} &= Y^p_{N+1} + e^p_{N+1} \end{align}

Now, $Y^p_{N+1}$ we have already calculated. So, if we want to bound $Y_{N+1}$ in an interval, say, 90% of the time, all we need to do is estimate consistently the $5^{th}$ and $95^{th}$ percentiles/quantiles of $e^p_{N+1}$, call them $e^5,e^{95}$, and the prediction interval will be $\left[Y^p_{N+1}+e^5,Y^p_{N+1}+e^{95} \right]$.

How to estimate the quantiles/percentiles of $e^p_{N+1}$? Well, we can write: \begin{align} e^p_{N+1} &= Y_{N+1}-Y^p_{N+1}\\ &= X_{N+1}\beta + \epsilon_{N+1} - X_{N+1}\hat{\beta}_{\text{OLS}}\\ &= X_{N+1}\left( \beta-\hat{\beta}_{\text{OLS}} \right) + \epsilon_{N+1} \end{align}

The strategy will be to sample (in a bootstrap kind of way) many times from $e^p_{N+1}$ and then calculate percentiles in the usual way. So, maybe we will sample 10,000 times from $e^p_{N+1}$, and then estimate the $5^{th}$ and $95^{th}$ percentiles as the $500^{th}$ and $9,500^{th}$ smallest members of the sample.

To draw on $X_{N+1}\left( \beta-\hat{\beta}_{\text{OLS}} \right)$, we can bootstrap errors (cases would be fine, too, but we are assuming iid errors anyway). So, on each bootstrap replication, you draw $N$ times with replacement from the variance-adjusted residuals (see next para) to get $\epsilon^*_i$, then make new $Y^*_i=X_i\hat{\beta}_{\text{OLS}}+\epsilon^*_i$, then run OLS on the new dataset, $\left(Y^*,X \right)$ to get this replication's $\beta^*_r$. At last, this replication's draw on $X_{N+1}\left( \beta-\hat{\beta}_{\text{OLS}} \right)$ is $X_{N+1}\left( \hat{\beta}_{\text{OLS}}-\beta^*_r \right)$

Given we are assuming iid $\epsilon$, the natural way to sample from the $\epsilon_{N+1}$ part of the equation is to use the residuals we have from the regression, $\left\{ e^*_1,e^*_2,\ldots,e^*_N \right\}$. Residuals have different and generally too small variances, so we will want to sample from $\left\{ s_1-\overline{s},s_2-\overline{s},\ldots,s_N-\overline{s} \right\}$, the variance-corrected residuals, where $s_i=e^*_i/\sqrt{(1-h_i)}$ and $h_i$ is the leverage of observation $i$.

And, finally, the algorithm for making a 90% prediction interval for $Y_{N+1}$, given that $X$ is $X_{N+1}$ is:

1. Make the prediction $Y^p_{N+1}=X_{N+1}\hat{\beta}_{\text{OLS}}$.
2. Make the variance-adjusted residuals, $\left\{ s_1-\overline{s},s_2-\overline{s},\ldots,s_N-\overline{s}\right\}$, where $s_i=e_i/\sqrt(1-h_{i})$.
3. For replications $r=1,2,\ldots,R$:
   • Draw $N$ times on the adjusted residuals to make bootstrap residuals $\left\{\epsilon^*_1,\epsilon^*_2,\ldots,\epsilon^*_N \right\}$
   • Generate bootstrap $Y^*=X\hat{\beta}_{\text{OLS}}+\epsilon^*$
   • Calculate bootstrap OLS estimator for this replication, $\beta^*_r=\left( X'X \right)^{-1}X'Y^*$
   • Obtain bootstrap residuals from this replication, $e^*_r=Y^*-X\beta^*_r$
   • Calculate bootstrap variance-adjusted residuals from this replication, $s^*-\overline{s^*}$
   • Draw one of the bootstrap variance-adjusted residuals from this replication, $\epsilon^*_{N+1,r}$
   • Calculate this replication's draw on $e^p_{N+1}$, $e^{p*}_r=X_{N+1}\left( \hat{\beta}_{\text{OLS}}-\beta^*_r \right)+\epsilon^*_{N+1,r}$
4. Find $5^{th}$ and $95^{th}$ percentiles of $e^p_{N+1}$, $e^5,e^{95}$
5. 90% prediction interval for $Y_{N+1}$ is $\left[Y^p_{N+1}+e^5,Y^p_{N+1}+e^{95} \right]$.

Here is R code:

# This script gives an example of the procedure to construct a prediction interval
# for a linear regression model using a bootstrap method.  The method is the one
# described in Section 6.3.3 of Davidson and Hinckley (1997),
# _Bootstrap Methods and Their Application_.


#rm(list=ls())
set.seed(12344321)
library(MASS)
library(Hmisc)

# Generate bivariate regression data
x <- runif(n=100,min=0,max=100)
y <- 1 + x + (rexp(n=100,rate=0.25)-4)

my.reg <- lm(y~x)
summary(my.reg)

# Predict y for x=78:
y.p <- coef(my.reg)["(Intercept)"] + coef(my.reg)["x"]*78
y.p

# Create adjusted residuals
leverage <- influence(my.reg)$hat
my.s.resid <- residuals(my.reg)/sqrt(1-leverage)
my.s.resid <- my.s.resid - mean(my.s.resid)


reg <- my.reg
s <- my.s.resid

the.replication <- function(reg,s,x_Np1=0){
  # Make bootstrap residuals
  ep.star <- sample(s,size=length(reg$residuals),replace=TRUE)
  
  # Make bootstrap Y
  y.star <- fitted(reg)+ep.star
  
  # Do bootstrap regression
  x <- model.frame(reg)[,2]
  bs.reg <- lm(y.star~x)
  
  # Create bootstrapped adjusted residuals
  bs.lev <- influence(bs.reg)$hat
  bs.s   <- residuals(bs.reg)/sqrt(1-bs.lev)
  bs.s   <- bs.s - mean(bs.s)
  
  # Calculate draw on prediction error
  xb.xb <- coef(my.reg)["(Intercept)"] - coef(bs.reg)["(Intercept)"] 
  xb.xb <- xb.xb + (coef(my.reg)["x"] - coef(bs.reg)["x"])*x_Np1
  return(unname(xb.xb + sample(bs.s,size=1)))
}

# Do bootstrap with 10,000 replications
ep.draws <- replicate(n=10000,the.replication(reg=my.reg,s=my.s.resid,x_Np1=78))

# Create prediction interval
y.p+quantile(ep.draws,probs=c(0.05,0.95))

# prediction interval using normal assumption
predict(my.reg,newdata=data.frame(x=78),interval="prediction",level=0.90)


# Quick and dirty Monte Carlo to see which prediction interval is better
# That is, what are the 5th and 95th percentiles of Y_{N+1}
# 
# To do it properly, I guess we would want to do the whole procedure above
# 10,000 times and then see what percentage of the time each prediction 
# interval covered Y_{N+1}

y.np1 <- 1 + 78 + (rexp(n=10000,rate=0.25)-4)
quantile(y.np1,probs=c(0.05,0.95))