Find all $n \in \mathbb{N}$ for which $(2^n + n) | (8^n + n)$. 
Find all $n \in \mathbb{N}$ for which $(2^n + n) | (8^n + n)$.

$n = 1, 2, 6$ are some solutions. Also, if the above holds then 
$$(2^n + n) | 2^n(2^n-1)(2^n+1)$$
and
$$(2^n + n)| n(2^n+1)(2^n-1)$$
I've tried using cases when $n$ is even or odd and have tried using modular arithmetic but am not able to proceed. Please help.
Thanks.
 A: Hint $\,\ 2^{\large n}\!+n\mid n+\color{#0a0}{8^{\large n}}\!\! \iff 2^{\large n}\!+n\mid n\color{#0a0}{-n^3}\ $ since
$\ {\rm mod}\,\ 2^{\large n}\!+n\!:\,\ \color{#c00}{2^{\large n}\equiv -n}\,\Rightarrow\, \color{#0a0}{8^{\large n}}\!= 2^{\large 3n}\!= (\color{#c00}{2^{\large n}})^{\large 3}\!\equiv (\color{#c00}{-n})^{\large 3}\!\equiv\color{#0a0}{-n^3}$
A: More generally,
if
$(a^n+n) |((a^3)^n+n)$,
then,
since
$(a^3)^n+n^3
=(a^n)^3+n^3
=(a^n+n)((a^2)^n-na^n+n^2)
$,
we have
$(a^3)^n+n
=(a^3)^n+n^3-n^3+n
=(a^n+n)((a^2)^n-na^n+n^2)-n^3+n
$
so
$(a^n+n)
| (n^3-n)
$.
Therefore
$a^n+n \le n^3-n$,
which bounds $n$
as a function of $a$.
In particular,
$a^n < n^3$,
so
$a < n^{3/n}$.
Since
$n^{3/n}
< e^{3/e}
< 3.1
$
for all $n$,
$4^{3/4}
< 3$,
and
$n^{3/n}
< 2$
for
$n \ge 10$,
we get these bounds on $n$:
If $a = 2$,
$n < 10$.
Trying
$\dfrac{8^n+n}{2^n+n}$,
the only integer values are,
according to Wolfy,
$(1, 3), (2, 11), (4, 205)$,
and
$(6, 3745)$.
If $a = 3$,
$n \le 3$.
Trying
$\dfrac{27^n+n}{3^n+n}$
we get,
again according to Wolfy,
the only solution is
$(1, 7)$.
There are no solutions for
$a \ge 4$.
