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 $p \equiv \pm 2 \bmod 5$.
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 $5^{(p-1)/2} = 5^{2^n k} \equiv -1 \bmod p$,
so $5^{mk} = (5^{2^n k})^a \equiv -1 \bmod p$
as well, which implies that
$5^m = 5^{2^n a} \not\equiv 1 \bmod p$.
On the other hand,
$2^m = (2^{2^n})^a \equiv (-1)^a = -1 \bmod p$,
so $p$ divides $2^m+1$, but does not divide $5^m-1$,
a contradiction.