Do we maximize explained sum of squares with OLS? I know that with OLS we minimize the sum of squared residuals but does that imply that we maximize SSE? From the following r-squared formula
$$ R^2 = \frac{SSE}{SST} = 1- \frac{SSR}{SST}$$
and the fact that SST = SSE + SSR, it really seems like it is true, but I am completely sure.

$SST = \sum_i (y_i - \bar y)^2$ is a constant that doesn't depend on your model, so doesn't change anything for optimization. If you look at the $R^2$ definition
$$
R^2 = 1 - \frac{SSE}{SST}
$$
you are left with $-SSE = - \sum_i (y_i - \hat y_i)^2$ that depends on the model. Maximizing the negative of the sum of squared errors is equivalent to minimizing squared error, which OLS does.
(Notice that I corrected the notation, you used $SSR$ in place of $SSE$ and vice versa.)
