Integers $n$ for which $|2^n + 5^n – 65|$ is a perfect square

Once you know that $n\geq 3$ must be even, let $ n = 2k$.

Then, apart from finitely many values of $k$,

$$5^{2k} < 5^{2k} + 2^{2k} - 65 < 5^{2k} + 2 \times 5^k + 1. $$

This bounds the expression strictly between 2 consecutive perfect squares, so it cannot be a perfect square.

So, we just need to check those finitely many values, which is left as an exercise to the reader.

NOTE: This is not a full answer and is intended to give a boost.

Taking modulo $4$, $$a^2-(0)^k\equiv 0\ (\text{mod}\ 4)$$ $$a\equiv0\ (\text{mod}\ 4)$$ Taking modulo $3$, $$a^2\equiv(-1)^n+(-1)^n-2\equiv 0\ (\text{mod}\ 3)$$ $$a\equiv0\ (\text{mod}\ 3)$$ Therefore, $a$ must be divisible by $12$.

Hope this part helps :)

EDIT: @CalvinLin has a wonderful (intended too I guess), so this will be left as intact, or if I should delete this, tell me.