If $a^{m}+1\mid a^{n}+1$ then prove that $m\mid n$. 
Let $a$ be an integer, $a\ge2$. If $a^{m}+1\mid a^{n}+1$ then prove that $m\mid n$.

Actually I know a similar proof which is, $a^{m}-1\mid a^{n}-1 \iff m\mid n$, but I can't prove this. I also need some examples of the question.
Can't seem to find any correlation between the two proofs.
I seem to not find examples where $a$ is something different from $2$ and taking $m=2$.
Please help. I think 4-5 examples might help me to see the proof.

$U_{k} = a^{k} + 1$ are terms of a Lucas sequence; hence, $U_{m} | U_{n}$ iff $m | n$.
