Question: Find $ n \in \mathbb{Z} $ such that $ \frac{8^n + n}{2^n + n} \in \mathbb{Z} $.
Here's what I have so far. Any hints would be appreciated.
Clearly $ n \geq 0 $ since otherwise $ 8^n + n < 2^n + n $. I use Wolfram to try several values of $ n $ and the ones that work seem to be only $ 0, 1, 2, 4, 6 $. Clearly $ n = 0 $ and $ n = 1 $ are easy to check, so we only need to consider $ n \geq 2 $. Suppose $ \frac{8^n + n}{2^n + n} = k \in \mathbb{Z} $, then $ (k - 1)n = 2^n(4^n - k) $. If $ k $ is even, then $ k - 1 $ is odd, so $ n $ must be even, but then $ n \not \mid 2^n $ since $ n < 2^n $. So $ k $ is odd. At this point I got stuck.
Using long division, we can write:
$$\frac{8^n+n}{2^n+n} = 4^n - 2^nn + n^2 - \frac{n^3-n}{2^n+n}$$
When $n\ge 10$, that last fraction cannot be an integer because the denominator becomes larger than the numerator. Thus, you only need to check values of $n<10$, which it appears you have done.