Solution:
Suppose $p=q$. Substituting we obtain $p^{2}(2-n)=1$ which is impossible because 1 is not divided by any prime. Hence necessarily $p$ and $q$ are different; since the equation is symmetric in $p$ and $q$ we may assume that $q>p$, that is $q \geq p+1$. Let us now write our equation as:
$$p^{2}-1=p q n-q^{2}=q(p n-q) .$$
This means that $p^{2}-1$ is a multiple of $q$ and therefore $q$ is a prime divisor of $p^{2}-1$. Hence
$$q \mid p^{2}-1=(p-1)(p+1) .$$
Now, since $q$ is a prime number, it must appear either in the factorization of $p-1$ or in that of $p+1$; in either case $q \leq p+1$. By the initial hypothesis we had $q \geq p+1$ and therefore we get $q=p+1$. Between two consecutive numbers one must necessarily be even and hence equal to 2 (the only even prime) and the other is necessarily 3 since 1 is not prime. Now let us check that the equation can be solved for $n$ by substituting $p=2$ and $q=3$:
$$4+9=6 n+1,$$
from which $n=2$. The only two solutions are therefore $(2,3,2)$ and $(3,2,2)$.