The equation is equivalent to $p(p+1) = (n-q)(n+q+1)$. Since the difference of the factors in the r.h.s. is greater than $1$, we must have $n-q < p$ and $n+q+1 > p+1$. As $p$ is a prime number, $p \mid n+q+1$. Let $n+q+1 = kp$, $k > 1$. Now the initial equation yields $p(p+1) = (kp-2q-1)kp$, which is equivalent to
$$2qk = (k+1)(pk - p - 1). \qquad (1)$$
As $k$ and $k+1$ are relatively prime, $2q \mid k+1$. Since $q$ is a prime number and $k > 1$, there are only two possibilities: $q = k+1$ or $2q = k+1$. In the first case, substituting $k$ to (1) gives $(p-2)(q-2) = 3$, implying $p=3$, $q=5$ (or the other way around). Then $n=6$ by the initial equation. In the second case similarly we get $(p-1)(q-1) = 1$, where the only possibility is $p=q=2$ and $n=3$.