Since $x^{2} - p x + q = 0$ has integral solution, then we may assume that
$$\begin{cases} p = a + b \\ q = a b \end{cases} \text{ with } a, b \in \mathbb{Z}^{+} .$$
Note that $q - p + 1 = a b - (a + b) + 1 = (a - 1)(b - 1) \geq 0$ then $q + 1 \geq p$.
By the same way on the second equation $x^{2} - q x + p = 0$, then we have $p + 1 \geq q$. Thus $p = q$ or $p + 1 = q$ or $q + 1 = p$. We consider three cases:

1. If $p = q$ then $x^{2} - p x + p = 0$ then we have
$$\Delta = p^{2} - 4p = k^{2} \Leftrightarrow (p - 2)^{2} - k^{2} = 4 .$$
Hence, $(p - 2 - k)(p - 2 + k) = 4$. Note that $(p - 2 - k) + (p - 2 + k) = 2(p - 2)$ then these numbers have the same parity, so we just need to consider $p - 2 - k = p - 2 + k = 2$ or $p = 4, k = 0$. In this case, we have one pair satisfies the given condition $(p, q) = (4, 4)$.

2. If $p + 1 = q$ then $x^{2} - p x + p + 1 = 0$, we have
$$\Delta = p^{2} - 4p - 4 = k^{2} \Leftrightarrow (p - 2)^{2} - k^{2} = 8 .$$
Hence $(p - 2 - k)(p - 2 + k) = 8$. Similarly, these numbers have the same parity then $p - 2 - k = 2$, $p - 2 + k = 4$ or $p = 5, k = 5$. In this case, we have one pair satisfies the given condition $(p, q) = (5, 6)$.

3. If $q + 1 = p$, similarly, we have $(p, q) = (6, 5)$.

Therefore, $(p, q) = (4, 4), (5, 6), (6, 5)$. $\square$