Write $p=2^m-1$ and $q=2^n-1$ where $m>n$ without loss of generality. Then $pq-1=(2^m-2^{m-n}-1)2^n$, and for this number to be a square, $n$ must be even, and $2^m-2^{m-n}-1$ must be a square, too. But if $m\ge n+2$, then $2^m-2^{m-n}-1\equiv 3\pmod 4$, so is not a square. Hence, we must have $m=n+1$, and then $2^m-2^{m-n}-1=2^m-3$. Clearly, this is not a square if $m$ is even. For $m$ odd this cannot be square, too, as in this case $2^m-3\equiv (-1)^m\equiv -1\pmod 3$.

Write $p=2^m-1$ and $q=2^n-1$ where $m>n$ without loss of generality. If $pq-1=(2^m-2^{m-n}-1)2^n$ is a square, then so is $2^m-2^{m-n}-1$. But then $m=n+1$, as $m\ge n+2$ implies $2^m-2^{m-n}-1\equiv 3\pmod 4$. Now, $2^m-2^{m-n}-1=2^m-3$, which clearly is not a square if $m$ is even, and which is not a square if $m$ is odd either as in this case $2^m-3\equiv (-1)^m\equiv -1\pmod 3$.