You're probably thinking of the two sample $t$ test because that's often the first place the $t$ distribution comes up. But really all a $t$ test means is that the reference distribution for the test statistic is a $t$ distribution. If $Z \sim \mathcal N(0,1)$ and $S^2 \sim \chi^2_d$ with $Z$ and $S^2$ independent, then
$$
\frac{Z}{\sqrt{S^2 / d}} \sim t_d
$$
by definition. I'm writing this out to emphasize that the $t$ distribution is just a name that was given to the distribution of this ratio because it comes up a lot, and anything of this form will have a $t$ distribution. For the two sample t test, this ratio appears because under the null the difference in means is a zero-mean Gaussian and the variance estimate for independent Gaussians is an independent $\chi^2$ (the independence can be shown via Basu's theorem which uses the fact that the standard variance estimate in a Gaussian sample is ancillary to the population mean, while the sample mean is complete and sufficient for that same quantity).
With linear regression we basically get the same thing. In vector form, $\hat \beta \sim \mathcal N(\beta, \sigma^2 (X^T X)^{-1})$. Let $S^2_j = (X^T X)^{-1}_{jj}$ and assume the predictors $X$ are non-random. If we knew $\sigma^2$ we'd have
$$
\frac{\hat \beta_j - 0}{\sigma S_j} \sim \mathcal N(0, 1)
$$
under the null $H_0 : \beta_j = 0$ so we'd actually have a Z test. But once we estimate $\sigma^2$ we end up with a $\chi^2$ random variable that, under our normality assumptions, turns out to be independent of our statistic $\hat \beta_j$ and then we get a $t$ distribution.
Here's the details of that: assume $y \sim \mathcal N(X\beta, \sigma^2 I)$. Letting $H = X(X^TX)^{-1}X^T$ be the hat matrix we have
$$
\|e\|^2 = \|(I-H)y\|^2 = y^T(I-H)y.
$$
$H$ is idempotent so we have the really nice result that
$$
y^T(I-H)y / \sigma^2 \sim \mathcal \chi_{n-p}^2(\delta)
$$
with non-centrality parameter $\delta = \beta^TX^T(I-H)X\beta = \beta^T(X^TX - X^T X)\beta = 0$, so actually this is a central $\chi^2$ with $n-p$ degrees of freedom (this is a special case of Cochran's theorem). I'm using $p$ to denote the number of columns of $X$, so if one column of $X$ gives the intercept then we'd have $p-1$ non-intercept predictors. Some authors use $p$ to be the number of non-intercept predictors so sometimes you might see something like $n-p-1$ in the degrees of freedom there, but it's all the same thing.
The result of this is that $E(e^Te / \sigma^2) = n-p$, so $\hat \sigma^2 := \frac{1}{n-p} e^T e$ works great as an estimator of $\sigma^2$.
This means that
$$
\frac{\hat \beta_j}{\hat \sigma S_j}= \frac{\hat \beta_j}{S_j\sqrt{e^Te / (n-p)}} = \frac{\hat \beta_j}{\sigma S_j\sqrt{\frac{e^Te}{\sigma^2(n-p)}}}
$$
is the ratio of a standard Gaussian to a chi squared divided by its degrees of freedom. To finish this, we need to show independence and we can use the following result:
Result: for $Z \sim \mathcal N_k(\mu, \Sigma)$ and matrices $A$ and $B$ in $\mathbb R^{l\times k}$ and $\mathbb R^{m\times k}$ respectively, $AZ$ and $BZ$ are independent if and only if $A\Sigma B^T = 0$ (this is exercise 58(b) in chapter 1 of Jun Shao's Mathematical Statistics).
We have $\hat \beta = (X^TX)^{-1}X^T y$ and $e = (I-H)y$ where $y \sim \mathcal N(X\beta, \sigma^2 I)$. This means
$$
(X^TX)^{-1}X^T \cdot \sigma^2 I \cdot (I-H)^T = \sigma^2 \left((X^TX)^{-1}X^T - (X^TX)^{-1}X^TX(X^TX)^{-1}X^T\right) = 0
$$
so $\hat \beta \perp e$, and therefore $\hat \beta \perp e^T e$.
The upshot is we now know
$$
\frac{\hat \beta_j}{\hat \sigma S_j} \sim t_{n-p}
$$
as desired (under all of the above assumptions).
Here's the proof of that result. Let $C = {A \choose B}$ be the $(l+m)\times k$ matrix formed by stacking $A$ on top of $B$. Then
$$
CZ = {AZ \choose BZ} \sim \mathcal N \left({A\mu \choose B\mu}, C\Sigma C^T \right)
$$
where
$$
C\Sigma C^T = {A \choose B} \Sigma \left(\begin{array}{cc} A^T & B^T \end{array}\right) = \left(\begin{array}{cc}A\Sigma A^T & A\Sigma B^T \\ B\Sigma A^T & B\Sigma B^T\end{array}\right).
$$
$CZ$ is a multivariate Gaussian and it is a well-known result that two components of a multivariate Gaussian are independent if and only if they are uncorrelated, so the condition $A\Sigma B^T = 0$ turns out to be exactly equivalent to the components $AZ$ and $BZ$ in $CZ$ being uncorrelated.
$\square$
