In maximum likleihood, we believe that the y-variable is conditionally normally distributed. So this means that errors are also normally distributed.
In ols regression, things seem to be more algebra/geometry driven. I am trying to fit a line of best fit between some points. I have done this in high school in my vector algebra class ... and there was never any mention of normal distribution in fitting a line of best fit.
So how come the errors in OLS are needed to be normally distributed?
In OLS, the errors do not have to have a normal distribution or even any particular distribution at all. All OLS does is solve the correspondence:
$$ \hat\beta_{OLS}\in\\ \underset{\beta=\left( \beta_0,\beta_1,\dots,\beta_p \right)}{\arg\min}\left\{ \overset{n}{\underset{i=1}{\sum}}\left( y_i-\left( \beta_0+\beta_1x_{i1}+\dots\beta_px_{ip} \right)\right)^2 \right\} $$
(I say that it is a correspondence instead of an equation because there does not have to be a unique solution $(\arg\min)$, such as if two features add up to a third.)
However, when you assume $iid$ Gaussian errors, the maximum likelihood estimate and OLS solution are equal.
That is the link.
As was mentioned by several commentators, ordinary least-squares does not require normal errors. Only when you want to say something about the sampling distribution of the OLS estimator in finite samples, normality needs to be assumed or else the estimates are not necessarily t-distributed.
To some extent, though, we're splitting hairs. The negative log-likelihood, which is being minimized to obtain $\hat{\beta}_{MLE}$, is algebraically identical to the least-squares objective function whose minimization leads to $\hat{\beta}_{OLS}$. If you further don't take the normal likelihood "too literal" but rather understand it as a second-order approximation of the "true" likelihood, you've pretty much eliminated all conceptual differences between OLS and (quasi)-MLE.
