For which $n$ does $2^n+1$ divide $10^n+1$? This came up in a discussion of numbers that divide their own binary representation (when interpreted as a decimal). The question I'm asking zooms in on a special case:

For which $n$ does $2^n+1$ divide $10^n+1$?

I computed the remainder after division of $10^n+1$ by $2^n+1$ for $1 \leq n \leq 300$, which led me to suppose that there are no such $n$.

In comments on the question, Chris has proven all the cases other than $ $$8|n$.
If $ $ $ $$n=8k$,
Assume $2^n+1$ divides $5^n-1$ (that is equivalent to the main question as you can see easily).
Take any prime $p$ that divides 
$2^{8k}+1$, then $p\neq 2,5$ obviously and
$$\Rightarrow p|2^{16k}-1 $$
Assume $2^m||16k$ for some $m\geq4$,
Because $p$ does not divide $2^{8k}-1$,
$$\Rightarrow 2^m|p-1$$
Because $2^{8k}+1$ divides $5^{8k}-1$,
 $p$ divides $5^{8k}-1$,
$\Rightarrow$ 5 is square in modulo $p$.
$\Rightarrow \left (\dfrac {p}{5} \right ) \left (\dfrac {5}{p} \right )=(-1)^{\frac {p-1}{2} \frac {5-1}{2} }=1$
$\Rightarrow \left (\dfrac {p}{5} \right )=1 \Rightarrow$ $ p$ is square in modulo 5.
$\Rightarrow p\equiv 1,-1 \pmod {5}$
$\Rightarrow $ all prime divisors of $2^n+1$ are of the form $5k+1$ or $5k-1$.
$\Rightarrow 2^n+1 $ should be $\equiv 1,-1 \pmod {5}$,but $ 2^n+1=2^{8k}+1\equiv 2 \pmod {5} \Rightarrow \Leftarrow $.
