Conjecture：
For any positive integer $m,n$, show that $(2^m-1)(3^n-1)$ is never a perfect square?
maybe this is old problem？ I find some $m,n$ the problem is right ,so How to prove it?
The conjecture was based on my solution to this following (different) problem :
Problem: >Find all non-negative integers $m$ and $n$, such that $(2^n-1) \cdot (3^n-1)=m^2$
Solution: Assume that $n > 0$ is a solution. First, note that $2 \mid n$, otherwise the exponent of $2$ in LHS in odd. From this (by lifting) we deduce that $n$ must be divisible by $3$ as well.
Let $n=6k$. Then $$ (2^{6k}-1)(3^{6k}-1) \equiv (2^k-1)(16^k-1) \equiv (2^k-1)^2 (2^k+1)(4^k+1) \pmod{31} $$ must be a quadratic residue. We claim this implies $5 \mid k$. Indeed, for $k \equiv 1,2,3,4 \pmod5$ we get $3 \cdot 5$, $5 \cdot 17$, $5 \cdot 13$, $17 \cdot 9$ are non-quadratic residues mod $31$, since of the primes here, only $5$ is a quadratic residue.
So let $n = 10c$. Then we have that $\left( (2^{10})^c - 1 \right)\left( (3^{10})^c - 1 \right)$ is a perfect square. But now by lifting the exponent on the prime $11$ we get a contradiction, because $\nu_{11}(2^{10}-1) = 1$ and $\nu_{11}(3^{10}-1) = 2$, thus the exponent of $11$ in the original is $3+2\nu_{11}(c)$.
Another partial solution, is to note that all squares of integers are 0 1 or 4 mod 8, $2^m-1$ is either 1,3 or 7 mod 8, and $3^n-1$ is either 0 or 2 mod 8.
$$1\cdot0\equiv0\bmod8; 1\cdot2\equiv2\bmod8; 3\cdot0\equiv0 \bmod8; 3\cdot2\equiv6\bmod8; 7\cdot0\equiv0\bmod8; 7\cdot2\equiv6\bmod8$$
None of these are 1 or 4 mod 8, so the only possibility left is 0 mod 8. when $3^n-1\equiv0\bmod8$ it's always one less than a square. so that part doesn't help. it then comes down can the square free parts for a square because a non primitive is the product of two or more squares. so either it's a primitive square in which case $2^m-1 =3^n-1$ which means there's a power of two that's a power of three, contradiction. So if it's not a primitive square it must factor into squares. since it takes $3^n-1\equiv0\bmod8$ for there to be a chance, and this factors as $(3^{n\over 2}+1)\cdot(3^{n\over2}-1)$ these must then half to combine with the factors of $2^m-1$ to make squares.