You want to produce a (positive) divisor of $p^mq^n$. By the Unique Factorization Theorem, aka the Fundamental Theorem of Arithmetic, this will be a number $d$ of the shape $p^aq^b$, where $0 \le a\le m$ and $0 \le b \le n$.
Imagine that we have a box that contains $m$ $p$'s, and next to it a box that contains $n$ $q$'s.
First we stop in front of the $p$-box, and decide how many $p$'s our divisor $d$ shall have. There are $m+1$ available options, namely $0, 1, \dots,m$.
Once we have decided on the number of $p$'s, move over to the $q$-box. For every choice of how many $p$'s the divisor $d$ shall have, there are $n+1$ ways to decide how many $q$'s the divisor $d$ shall have. Thus the total number of choices is $(m+1)(n+1)$.
Comment: Let $N=p_1^{m_1}p_2^{m_2}\cdots p_k^{m_k}$, where the $p_i$ are distinct primes. Using basically the same argument, we can show that $N$ has $(p_1+1)(p_2+1)\cdots(p_k+1)$ positive divisors.