A group of order $p^2q^2$ is never simple
We may as well assume $p < q$. The number of Sylow $q$-subgroups is $1$ mod $q$ and divides $p^2$. So it is $1, p$, or $p^2$. We win if it's $1$ and it can't be $p$, so suppose it's $p^2$. But now $q \mid p^2 - 1$, so $q \mid p+1$ or $q \mid p-1$.
Thus $p = 2$ and $q = 3$. The case of order $36$ has been proved in an earlier question:
No group of order 36 is simple
Using a little more group theory allows us to prove something stronger (and avoid the reduction to $|G|=36$):
A group of order $p^2q^2$ has either a normal Sylow $p$-group or normal Sylow $q$-group.
For assume that $p<q$, then there are either $1$ or $p^2$ Sylow $q$-groups in $G$.
If there is $1$, it is normal, and we are done.
If there is $p^2$, then the Sylow $q$-groups are self-normalizing. But any group of order $q^2$ is abelian, so Burnside's transfer theorem implies the Sylow $p$ group is normal.