Showing that a Sylow subgroup of a group of order $p^{2}q$ is normal.

Let $N_q$ be the number of Sylow $q$-subgroups. Then $N_q = 1$ or $p$ or $p^2$.

Suppose $N_q = p$. Since $p \equiv 1$ (mod $q$), $p > q$. Similarly if the number of Sylow $p$-subgroup is $q$, $q > p$. This is a contradiction. Hence there is only one Sylow $p$-subgroup.

Suppose $N_q = p^2$. Then the number of elements of order $q$ is $p^2(q - 1)$. Hence the number of elements of order not equal to $q$ is $p^2q - p^2(q - 1) = p^2$. Hence there is only one Sylow $p$-subgroup.

As you noted $n_q=1,p,p^2$. I assume $p<q$. If $n_q=p$ then $n_q=1+kq=p\Longrightarrow p>q$ which is a contradiction. If $n_q=p^2$, let $Q_1, Q_2$ be two distinct $q-$ sylow subgroup. They are both cyclic of order $q$. $Q_1\cap Q_2\le Q_1$ then $|Q_1\cap Q_2|\bigg|q$ so $|Q_1\cap Q_2|=1$ or $=q$. if $|Q_1\cap Q_2|=q$ then $Q_1\cap Q_2=Q_1$ which is wrong so $|Q_1\cap Q_2|=1$. Hence, every two $q-$ sylow subgroup has $1$ in common. Now enumerate all non-trivial element of the group which lie in these $p^2$ subgroups. It is as @Makoto pointed. It is exactly the order of a $p$ sylow subgroup of $G$. Since there is not any member in shared among $p-$ sylow subgroups and $q-$ sylow subgroups so the $p-$ sylow is normal in the group.