Solution:

Rearranging the equation, $2 q n(p+1) = (n+2)(2 p q + p + q + 1)$. The left hand side is even, so either $n+2$ or $p+q+1$ is even, so either $p=2$ or $q=2$ since $p$ and $q$ are prime, or $n$ is even.

If $p=2$, $6 q n = (n+2)(5 q + 3)$, so $(q-3)(n-10) = 36$. Considering the divisors of $36$ for which $q$ is prime, we find the possible solutions $(p, q, n)$ in this case are $(2,5,28)$ and $(2,7,19)$ (both of which satisfy the equation).

If $q=2$, $4 n(p+1) = (n+2)(5 p + 3)$, so $n = p n + 10 p + 6$, a contradiction since $n < p n$, so there is no solution with $q=2$.

Finally, suppose that $n=2 k$ is even. We may suppose also that $p$ and $q$ are odd primes. The equation becomes $2 k q(p+1) = (k+1)(2 p q + p + q + 1)$. The left hand side is even and $2 p q + p + q + 1$ is odd, so $k+1$ is even, so $k = 2 \ell + 1$ is odd. We now have
$$q(p+1)(2 \ell + 1) = (\ell + 1)(2 p q + p + q + 1)$$
or equivalently
$$\ell q(p+1) = (\ell + 1)(p q + p + 1)$$
Note that $q \mid p q + p + 1$ if and only if $q \mid p+1$. Furthermore, because $(p, p+1) = 1$ and $q$ is prime, $(p+1, p q + p + 1) = (p+1, p q) = (p+1, q) > 1$ if and only if $q \mid p+1$.

Since $(\ell, \ell+1) = 1$, we see that, if $q \nmid p+1$, then $\ell = p q + p + 1$ and $\ell + 1 = q(p+1)$, so $q = p+2$ (and $(p, p+2, 2(2 p^{2} + 6 p + 3))$ satisfies the original equation). In the contrary case, suppose $p+1 = r q$, so $\ell(p+1) = (\ell+1)(p+r)$, a contradiction since $\ell < \ell+1$ and $p+1 \leq p+r$.

Thus the possible values of $q-p$ are $2, 3$ and $5$.

Solution 2:

Subtracting $2$ and multiplying by $-1$, the condition is equivalent to
$$\frac{1}{p+1} - \frac{1}{q} = \frac{4}{n+2}$$
Thus $q > p+1$. Rearranging,
$$q-p-1 = \frac{4(p+1)q}{n+2}$$
The expression on the right is a positive integer, and $q$ must cancel into $n+2$ else $q$ would divide $p+1 < q$. Let $(n+2)/q = u$ a positive integer.

So
$$q-p-1 = \frac{4(p+1)}{u}$$
$$u q - u(p+1) = 4(p+1)$$
so $p+1$ divides $u q$. However, $q$ is prime and $p+1 < q$, therefore $p+1$ divides $u$. Let $v$ be the integer $u/(p+1)$. Now
$$q-p = 1 + \frac{4}{v} \in \{2,3,5\}$$
All three cases can occur, where $(p, q, n)$ is $(3,5,78)$, $(2,5,28)$ or $(2,7,19)$. Note that all pairs of twin primes $q = p+2$ yield solutions $(p, p+2, 2(2 p^{2} + 6 p + 3))$.

Solution 3:

Subtract $2$ from both sides to get
$$\frac{1}{p+1} - \frac{1}{q} = \frac{4}{n+2}$$
From this, since $n$ is positive, we have that $q > p+1$. Therefore $q$ and $p+1$ are coprime, since $q$ is prime.

Group the terms on the LHS to get
$$\frac{q-p-1}{q(p+1)} = \frac{4}{n+2}$$
Now $(q, q-p-1) = (q, p+1) = 1$ and $(p+1, q-p-1) = (p+1, q) = 1$ so the fraction on the left is in lowest terms. Therefore the numerator must divide the numerator on the right, which is $4$. Since $q-p-1$ is positive, it must be $1, 2$ or $4$, so that $q-p$ must be $2, 3$ or $5$. All of these can be attained, by $(p, q, n) = (3,5,78)$, $(2,5,28)$ and $(2,7,19)$ respectively.