Proving $11$ never divides $64^n +22^n -2$

The question tells you to use the Division Theorem, here is my attempt:

Every integer can be expressed in the form $7q+r$ where $r$ is one of $0,1,2,3,4,5$ or $6$ and $q$ is an integer $\geq0$.

$n=7q+r$

$n^2=(7q+r)^2=49q^2+14rq+r^2$

$n^2=7(7q^2+2rq)+r^2$

$n^2+4=7(7q^2+2rq)+r^2+4$

$7(7q^2+2rq)$ is either divisible by $7$, or it is $0$ (when $q=0$), so it is $r^2+4$ we are concerned with.

Assume that $r^2+4$ is divisible by 7. Then $r^2+4=7k$ for some integer $k$.

This is the original problem we were faced with, except whereas $n$ could be any integer, $r$ is constrained to be one of $0,1,2,3,4,5$ or $6$.

Through trial and error, we see that no valid value of $r$ satisfies $r^2+4=7k$ so we have proved our theorem by contradiction.

I'm pretty sure this is either wrong somewhere or at the very least not the proof that the question intended. Any help would be appreciated.

since we have $$n\equiv 0,1,2,3,4,5,6\mod 7$$ we get $$n^2\equiv 0,1,2,4\mod 7$$ therefore $$n^2+4\equiv 1,4,5,6\mod 7$$

Your proof is correct (fleablood's comment). Let me rephrase a bit:

$A(n):= n^2 +4$; $B(r,q) = 7 q^2 + 2rq;$

$A(n) = 7B(r,q) + (r^2 +4)$.

Fairly simple to show that:

$A(n)$ is divisible by $7 \iff $

$(r^2 +4)$ is divisible $ n.$

By inspection

$(r^2 +4) , r = 0,1,2,3,4,5,6,$

is not divisible by $7$.

$\Rightarrow$: $A(n)$ is not divisible by $7$.