I saw this homework many times, but always asks in the statament that $p^2 \not\equiv 1$ (mod $q$) and $q \not\equiv 1$ (mod $p$).
But today in a book text of Galois theory I Saw a similar example that the statament just asks that $q$ doesn't divides $p^2$.
Just using this fact i Tried to prove that exists just one subgroup with order $p^2$ and just one with order $q$
By the first Sylow theorem groups with such order exists, and by the third theorem the number of this groups are of the form $n_{p^2}=1+kp^2, k\in \mathbb{N}$ and $n_{q}=1+k'q, k'\in \mathbb{N}$ with boths dividing the order of $G$.
$n_{p^2}$ divides $p^2$ just for $k=0$ and by the hyphotesis $q\nmid p^2$ how Could I conclude that there is no $k\neq 0$ such that $n_{p^2}\nmid q$?
Analogously how to use this to prove $n_{q}\nmid p^2$?
@Eduardo Silva: $q$ doesn't divide $p^2$ is not enough (in fact, it is equivalent to $p \neq q$), and so Iulu example works as a counterexample. I guess the only thing you can say with your hypoteses is that there is only one Sylow p-Subgroup (since $p>q$, $n_p$ must be $1$, for the next available value is $p+1$, which cannot divide $q$). Are you sure there is no other assumption?