I am trying to show that any group of order $p^2q$ has a normal Sylow subgroup where $p$ and $q$ are distinct primes.

In the case $p>q$ I have no problem.. By Sylow $n\_p|q$, so $n\_p$ is either $1$ or $q$. But we also have $n\_p\equiv1\mod p$ which rules out $q$. So in the case $p>q$, $n\_p=1$ and the unique p-Sylow subgroup is normal.

In the case $p<q$ however, I run into a problem.. Again we have $n\_q|p^2$ (so $n\_q\in{1,p,p^2}$). Again, the condition $n\_q\equiv1\mod q$ rules out $p$.
Now I am attempting to rule out the case $n\_q=p^2$: Assume $n\_q=p^2$. Then $p^2\equiv1\mod q\implies (p+1)(p-1)\equiv0\mod q\implies q$ must divide $(p+1)$ since $p<q$ and $q$ prime. But since $p<q$ and $q|(p+1)$ we see $q=p+1$. For primes this only happens when $p=2, q=3$.

Does this mean I need to check groups of order $2^2\cdot3=12$ or did I miss something along the way that lets me conclude $n\_q\neq p^2$ and thus that $n\_q=1$.