Find the solutions to the equation $(2^m-1) = (2^n-1)k^2$ where $m,n,k$ are positive integers.
One solution is $m = n$.
Since $2^n-1 \mid 2^m-1$, it follows that $n \mid m$ because $\gcd(2^m-1,2^n-1) = 2^{\gcd(m,n)}-1$. Let $m = nd$ for some positive integer $d$. Then we have $$\dfrac{2^m-1}{2^n-1} = \dfrac{2^{nd}-1}{2^n-1} = 2^{(d-1)n}+2^{(d-2)n}+\cdots+2^n+1 = k^2.$$ Therefore we can write $k^2 = 1u \ldots u_2$ where $u = \underbrace{00 \ldots 0}_{n-1 \text{ }\text{zeros}}1$ with $u$ appearing $d-1$ times. How can we continue from here?
If $d:=m/n$ is even, let $d=2r$. Then $$k^2=\frac{2^m-1}{2^n-1}=\frac{2^{nr}-1}{2^n-1}(2^{nr}+1).$$ Since $k^2$ factors into two coprime positive integers, both of the two factors must be perfect square. Then $2^{nr}+1=z^2$ leads to $(z-1)(z+1)=2^{nr}$, so $z-1$ and $z+1$ are both powers of $2$. This only happens when $z=3,nr=3$. It remains to check that $k\not\in\mathbb{Z}$ when $n=1,r=3$ and $k=3$ when $n=3,r=1$.
Therefore, the only solution is $(m,n,k)=(6,3,3)$ given that $d$ is even.
If $d$ is odd, there are two cases.
Case 1: For $n=2l$, consider the Pell equation $$x^2-(2^{2l}-1)y^2=1\tag{1}\label{1}$$ It is easy to see that $(2^l,1)$ is the fundamental solution of \eqref{1}.
The equation $2^{nd}-1=(2^n-1)k^2$ implies that $(2^{ld},k)$ is also a solution of \eqref{1}. So we have $$2^{ld}+k\sqrt{2^{2l}-1}=\biggl(2^l+\sqrt{2^{2l}-1}\biggr)^t$$ for integer $t>0$. Take the reciprocal to get $$2^{ld}-k\sqrt{2^{2l}-1}=\biggl(2^l-\sqrt{2^{2l}-1}\biggr)^t.$$ Let $s=\lfloor t/2\rfloor$, then \begin{align} 2^{ld}&=\frac{1}{2}\Biggl(\biggl(2^l+\sqrt{2^{2l}-1}\biggr)^t+\biggl(2^l-\sqrt{2^{2l}-1}\biggr)^t\Biggr)\\ &=\sum_{i=0}^{s}\binom{t}{2i}2^{(t-2i)l}(2^{2l}-1)^i\\ &=\binom{t}{2s}2^{(t-2s)l}(2^{2l}-1)^{s}+2^{2l}\sum_{i=0}^{s-1}\binom{t}{2i}2^{(t-2i-2)l}(2^{2l}-1)^i.\\ \end{align} Consider the exponent of $2$, we have $$\nu_2\biggl(\binom{t}{2s}2^{(t-2s)l}(2^{2l}-1)^{s}\biggr)=(t-2s)l\leqslant l<2l\leqslant\nu_2\Biggl(2^{2l}\sum_{i=0}^{s-1}\binom{t}{2i}2^{(t-2i-2)l}(2^{2l}-1)^i\Biggr)$$ So $0<ld=\nu_2(2^{ld})\leqslant l$, which implies $d=1$.
Case 2: For $n=2l-1$, consider the Pell equation $$x^2-(2^{2l}-2)y^2=1\tag{2}\label{2}$$ We claim that $(2^{2l}-1,2^l)$ is the fundamental solution of \eqref{2}.
Obviously, $(2^{2l}-1,2^l)$ is a solution of \eqref{2}. Let $(u,v)$ be the fundamental solution, then $$2^{2l}-1+2^l\sqrt{2^{2l}-2}=\biggl(u+v\sqrt{2^{2l}-2}\biggr)^e$$ for integer $e>0$ and we have $2^{2l}-1\geqslant u^e$. Since it is easy to see that $u>2^l$, we get $e=1$ thus $(2^{2l}-1,2^l)=(u,v)$ is the fundamental solution as we claimed.
Because $2^{nd}-1=(2^n-1)k^2=(2^{2l-1}-1)k^2$ implies $$(2^{nd+1}-1)^2-(2^{2l}-2)k^22^{nd+1}=(2^{nd+1}-1)^2-(2^{nd}-1)2^{nd+2}=1,$$ i.e. $(2^{nd+1}-1,k2^{(nd+1)/2})$ is also a solution of \eqref{2}, we have $$2^{nd+1}-1+k2^{(nd+1)/2}\sqrt{2^{2l}-2}=\biggl(2^{2l}-1+2^l\sqrt{2^{2l}-2}\biggr)^t$$ for integer $t>0$. The same as above, we transform it to get \begin{align} 2^{nd+1}&=1+\frac{1}{2}\left(\left(2^{2l}-1+2^l\sqrt{2^{2l}-2}\right)^t+\left(2^{2l}-1-2^l\sqrt{2^{2l}-2}\right)^t\right)\\ &=1+\sum_{i=0}^{\lfloor t/2\rfloor}\binom{t}{2i}(2^{2l}-1)^{t-2i}2^{2li}(2^{2l}-2)^i\\ &=1+(2^{2l}-1)^t+\sum_{i=1}^{\lfloor t/2\rfloor}\binom{t}{2i}(2^{2l}-1)^{t-2i}2^{(2l+1)i}(2^{2l-1}-1)^i\\ &=1+(-1)^t+(-1)^{t-1}t2^{2l}+2^{2l+1}R. \end{align} Note that $$\nu_2\bigl(1+(-1)^t+(-1)^{t-1}t2^{2l}\bigr)=\begin{cases} 1, & \text{$t$ even}\\ 2l, & \text{$t$ odd} \end{cases}\leqslant 2l<2l+1\leqslant\nu_2(2^{2l+1}R).$$ So $1<nd+1=\nu_2(2^{nd+1})\leqslant2l=n+1$, which also implies $d=1$.
Therefore, we always get $d=1$ given that $d$ is odd.
From above, $(m,n,k)=(6,3,3),(n,n,1)$ are the only solutions of the equation.
Background
W. Ljunggren had proved a theorem that for $n>2$, the equation $$y^2=\frac{x^n-1}{x-1}$$ has no solutions in integers for $|x|>1$, except $n=4,x=7$ and $n=5,x=3$.
This theorem will expel most cases of your equation and you can find a proof in this post. It is a short proof based on the results of T. Nagell and K. Mahler.
For a self-contained answer to your question, I borrowed Ljunggren's idea to transform the equation into a Pell(-like) equation, and made use of the special setting of $x$ to finish my proof without the help of other results. However, the knowledge of Pell equation is required in advance.