Am I going about testing for convergence correctly for this example? $$\sum_{n=1}^\infty \frac{(2+4^{-n})}{n^2}$$

Would I be correct in thinking that this is the comparison test and need to use something like $\cfrac{3}{n^2}$ ? This is a much more advanced fraction than in my notes and am cautious of the power of -n on the numerator
Your solution is good. You could even make it closer if you notice that, $\forall n \geq 1$, $4^{-n} \leq \frac 14$ making
$$\sum_{n=1}^\infty \frac{(2+4^{-n})}{n^2} \lt \frac 94\sum_{n=1}^\infty \frac{1}{n^2}$$
Still simpler: asymptotic equivalence. The numerator of the  general term is equivalent to $2$, so
$$\frac{(2+4^{-n})}{n^2}\sim_\infty\frac{2}{n^2},$$
which converges.