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

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 $.

Pleased as I am with Merdanov's great answer, I would like to reformulate it, including the bit supplied by Chris, so as to make it a bit more concise and compatible with my way of looking at these things.

First the part done by Chris (actually, he went further, but I am only using what I need):

Assume $2^n+1$ divides $10^n+1$. We prove that $n$ must be a multiple of $4$. Indeed, this was shown by Chris in the comments, since if $n$ were odd, we get $2^n+1 \equiv 0 \pmod{3}$ whereas $10^n+1 \equiv 2 \pmod{3}$, and if $n$ were $2\pmod{4}$ we would have $2^n+1 \equiv 0 \pmod{5}$ whereas $10^n+1 \equiv 1 \pmod{5}$.

As Chris suggested, we could probably go on with this line of reasoning; see the comments on Merdanov's answer for more on this.

So now to Merdanov's answer. I use some (very) basic group theory in my reformulation, which is already implicit in what Merdanov wrote.

First observe that $2^n+1$ must also divide $-(10^n+1)+5^n (2^n+1)$, which equals $5^n-1$. Let $p$ be a prime dividing $2^n+1$, then of course $p \mid 5^n-1$ as well, and $p \neq 2,5$. Since $p$ divides $2^n+1$ but not $2^n-1$, we have that $2n$ is a multiple of the order of $2$ in the cyclic group $C:=(\mathbb{Z}/p\mathbb{Z})^\times$, whereas $n$ is not. Let $2^m$ be the highest power of $2$ dividing $2n$. Then $2^m$ divides the order of $2$ in $C$, hence also divides the order of $C$.

Since $p \mid 5^n-1$, we have that $n$ is a multiple of the order of $5$ in $C$. Hence $2^{m-1}$ is the highest power of $2$ dividing the order of $5$ in $(\mathbb{Z}/p\mathbb{Z})^\times$, which means $5$ is a square in $(\mathbb{Z}/p\mathbb{Z})^\times$. By quadratic reciprocity, $p$ is a square modulo $5$, hence $p \equiv \pm 1 \pmod{5}$. Since $p$ was arbitrary, $2^n+1$ is a product of primes of the form $\pm 1 \pmod{5}$. Since we had already shown that $4 \mid n$, we get $2^n+1\equiv 1+1 \equiv 2 \pmod{5}$, which is a contradiction.

This does suggest that "the numbers worked in our favour". If $10$ hadn't been a multiple of $2$ (if this slight abuse of language is permitted), then we couldn't have played the orders of the two elements of $(\mathbb{Z}/p\mathbb{Z})^\times$ off against each other the way we did. Problems of the type where $a^n-1$ is supposed to divide $b^n+1$ would seem to be even more difficult to accomodate. But that would be a topic for some other time...