Find all $m$ such $2^m+1\mid5^m-1$ 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
$$16^p+1\mid625^p-1.$$
 A: 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.
A: 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.
