So recently I've received a problem (title) and I have no idea how to solve it. By easy observation, $(a,b)=(m,m),(6,3),(3,6)$ are the cases I've found, and no others.
So I try to let $kA^2=2^a-1$ and $kB^2=2^b-1$, where $k = \gcd(2^a-1,2^b-1)=2^{\gcd(a,b)}-1$. Then, another equation appears:$$\dfrac{2^a-1}{2^{\gcd(a,b)}-1}=A^2$$ the other one is similar. This expression is a bit ugly so I let $\gcd(a,b)=x$ , $a=xy$. The equation then becomes: $$\dfrac{2^{xy}-1}{2^x-1}=A^2$$ which is much nicer. The solutions I have found for now (which is according to the (a,b) solutions above) are $(x,y)=(x,1),(3,2)$. Also, by modulus and some algebra, if there are more solutions we just found, then $x\ge3$ and $y\ge 5$. Well that's it. I can't go further more. Please give me some idea or even tips or solutions. Also, no computer is allowed. Thank you.