Edit: Dietrich Burde kindly gave me a hint but I, in my inexperience, am not able to understand the logic. I've tried but failed. So, right now my problem is understanding him. So, if you can explain the logic to me, I would be grateful.
For which positive integer $n$, is $\frac{2^n-1}{3}$ a factor of $4m^2+1$ for some integer $m$? So, I've gotten that $n=2^q,\ q\in\mathbb{Z}_+$ is one solution, and I'm pretty certain that it is the only solution, but I don't know how to prove this (correct or wrong).
If you're interested on how I got to this point, well here's briefly what I've done: $n$ must be an even integer, as $3\mid 2^n-1$ as $2^{2k}-1=4^k-1\equiv 1^k-1\equiv 0\ (mod\ 3)$
$n=2$ is a quite trivial solution.
$\frac{2^n-1}{3}$ must be a factor of $4m^2+1$, which gives us this (Diophantine) equation: $$4m^2+1=\frac{d(2^n-1)}{3}$$ For $d=1$, we get $4m^2=\frac{2^n-4}{3}$ and because $n$ must be even, we get: $$4m^2=\frac{4^k-4}{3}$$ $$m^2=\frac{4^{k-1}-1}{3}$$ $$3m^2=(2^{k-1})^2-1^2$$ $$3m^2=(2^{k-1}-1)(2^{k-1}+1)$$ From this you can quite easily get $k=1, m=\pm 1$ as a solution so $n=4$.
But what about for other values of $d$?
Well, we have $n=2,4=2^1, 2^2$. Maybe it works for all $n=2^q$. I proved this with induction using the fact that $2^{(2^{l+1})}=(2^2)^{2^l}$.
But is there other solutions? I don't know. I feel like there is no other ones because I haven't been able to find anything for the integers in between.
Hint: Every divisor $d$ of $4m^2+1$ satisfies $d\equiv 1\bmod 4$. And $\frac{2^n-1}{3}$ is an integer if and only if $n$ is even. Suppose that $n=2k$ is not a power of $2$. Then $\frac{2^n-1}{3}$ has a prime divisor $p$ with $p\equiv 3\bmod 4$, a contradiction.