A group of order $p^2q$ will be abelian

Since $ q\nmid p^{2}-1$ therefore $n_{q} = 1+kq \neq p,\: p^{2}$. So we also have a normal $q$ -Sylow subgroup. So

• $G \cong \mathbb{Z}_{p^2} \times \mathbb{Z}_{q}$

• $G \cong \mathbb{Z}_{p} \times \mathbb{Z}_{p} \times \mathbb{Z}_{q}$.

Hence $G$ is abelian.

@Babak, your problem is a special case of the following.

A positive integer $n=p_1^{a_1}. .. p_t^{a_t}$, $p_i$ distinct, is said to have good factorization if and only if $p_i^{k} \not\equiv 1$ mod $p_j$ for all integers $i$, $j$ and $k$, where $1 \leqslant k \leqslant a_i$.

Theorem The groups of order n are all abelian if and only if n is cube-free and has good factorization.

See for example the nice paper Nilpotent and Solvable Numbers by J. Pakianathan and K. Shankar