Suppose $(2^m-1)(2^n-1)-1=2^{m+n}-2^m-2^n$ is a square. Obviously we can't have $m=n$, so the largest power of 2 dividing this is $2^{\min(m,n)}$. If neither of our Mersenne primes is 3, then that exponent is an odd number, which can't possibly be the largest power of 2 dividing a square.

So one of our primes is 3 and we have $2^{n+2}-2^n-4$ where $n$ is an odd prime. This is 4 times $2^n-2^{n-2}-1$. Unless $n=3$ this is $-1$ mod 4 and therefore can't be a square.

So the only remaining case to check is $m=2,n=3$ yielding $3.7-1=20$ which happens not to be a square.

Suppose $(2^m-1)(2^n-1)-1=2^{m+n}-2^m-2^n$ is a square. Obviously we can't have $m=n$, so the largest power of 2 dividing this is $2^{\min(m,n)}$. If neither of our Mersenne primes is 3, then that exponent is an odd number, which can't possibly be the largest power of 2 dividing a square.

So one of our primes, say $2^m-1$, is 3 and we have $2^{n+2}-2^n-4$ where $n$ is an odd prime. This is 4 times $2^n-2^{n-2}-1$. Unless $n=3$ this is $-1$ mod 4 and therefore can't be a square.

So the only remaining case to check is $m=2,n=3$ yielding $3\cdot7-1=20$ which happens not to be a square.