$(2^m -1)(2^n-1)$ divides $(2^{mn} -1)$ if and only if $\gcd(m,n) = 1$. If $\gcd(m,n) = 1$ then $(2^m-1)(2^n-1)$ divides $2^{mn} - 1$ because each of $2^m-1$, $2^n-1$ divide $2^{mn}-1$ and $\gcd(2^m-1, 2^n-1) = 2^{\gcd(m,n)}-1 = 1$. How about the converse? If $\gcd(m,n) > 1$, can one show that $(2^m-1)(2^n-1)$ does not divide $(2^{mn}-1)$?

The following generalization can be proved by a straightforward adaptation of the proof I gave above:
Given integer $a \geq 2$, positive integers $m,n$, and $k$ a positive integer that is a common divisor of $m$, $n$, then
$(a^m-1)(a^n-1)/(a^k-1)$ divides $a^{mn}-1$ if and only if $k = \gcd(m,n)$.
