The problem comes from a problem I encountered when I wrote the article
Find all positive integer $m$ such $$2^{m}+1\mid5^m-1$$ it seem there no solution. I think it might be necessary to use quadratic reciprocity knowledge to solve this problem. If $m$ is odd then $2^m+1$ is divisible by 3 but $5^m-1$ is not. so $m$ be even, take $m=2n$, then $$4^n+1\mid25^n-1.$$ if $n$ is odd,then $4^n+1$ is divisible by $5$, but $25^n-1$ is not so $n$ is even,take $n=2p$, we have $$16^p+1\mid625^p-1.$$
Here is a proof.
Theorem. $2^m+1$ never divides $5^m-1$.
Assume that there is some $m$ such that $2^m+1$ divides $5^m-1$. We already know that $m$ must be divisible by $4$. Let $m = 2^n a$ with an odd integer $a$ and $n \ge 2$. The $n$th Fermat number $$F_n = 2^{2^n} + 1$$ is congruent to $2$ mod $5$ (this uses $n \ge 2$), so it has a prime divisor $p$ such that $$p \equiv \pm 2 \pmod 5.$$ We know that $p-1 = 2^{n+1}k$ for some integer $k$. Since $\left(\frac{5}{p}\right) = \left(\frac{p}{5}\right) = -1$, we have that $$5^{2^n k} = 5^{(p-1)/2} \equiv -1 \pmod p,$$ so $$5^{mk} = (5^{2^n k})^a \equiv -1 \pmod p$$ as well. In particular, $$5^m \not\equiv 1 \pmod p.$$ On the other hand, $$2^m = (2^{2^n})^a \equiv (-1)^a = -1 \pmod p.$$ Thus $p$ divides $2^m+1$, but does not divide $5^m-1$, a contradiction.
There might well be a very elementary argument for this, but in the spirit of taking a hammer to a fly, one can prove that the number of $m$ such that $$ 2^m+1 \mid 5^m-1 $$ is finite by invoking a theorem of Bugeaud, Corvaja and Zannier [Math. Z. 2003] which implies, in this context that, given $\epsilon > 0$, $$ \gcd (4^m-1, 5^m-1) \leq e^{\epsilon m} $$ for suitably large $m$. Schmidt's Subspace Theorem is used here, so the result is ineffective.
