There are given different odd prime numbers $\displaystyle p$ and $\displaystyle q$. Prove that number $\displaystyle 2^{pq} -1$ has at least 3 different prime factors.

From what I found in an algebra book:

$\displaystyle a^n\,-\,b^n \;=\;(a\,-\,b)\,(a^{n-1}\,+\,a^{n-2}b\,+\,a^{n-3}b^2\,+\,\cdots\.+\,ab^{n-2}\,+\,b^{n-1})$

So maybe we can factorize this in two ways:

. . $\displaystyle (2^p)^q \,-\,1^q \;=\;(2^p\,-\,1)\left([2^p]^{q-1}\,+\,[2^p]^{q-2}\,+\,\cdots\,+[2^p]\,+\,1\right)$

. . $\displaystyle (2^q)^p \,-\,1^p \;=\;(2^q\,-\,1)\left([2^q]^{p-1}\,+\,[2^q]^{p-2} \,+\,\cdots\,+\,[2^q]\,+\,1\right)$

But I still don't know what to do next and if my steps were OK... Could you please help me?