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 $$5^{(p-1)/2} = 5^{2^n k} \equiv -1 \pmod p,$$ so $$5^{mk} = (5^{2^n k})^a \equiv -1 \pmod p$$ as well, which implies that $$5^m = 5^{2^n a} \not\equiv 1 \pmod p.$$ 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.

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.

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.

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 $$5^{(p-1)/2} = 5^{2^n k} \equiv -1 \pmod p,$$ so $$5^{mk} = (5^{2^n k})^a \equiv -1 \pmod p$$ as well, which implies that $$5^m = 5^{2^n a} \not\equiv 1 \pmod p.$$ 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.