If one of the numbers is $2$, then $p = q = 2$, and we can assume $p, q \ge 3$. Since $(p, p^2 + 8) = (q, q^2 + 8) = 1$, we have: $q \mid p^2 + 8$ and $p \mid q^2 + 8 \Rightarrow pq \mid (p^2 + 8)(q^2 + 8) \Rightarrow pq \mid 8(p^2 + q^2 + 8) \Rightarrow pq \mid (p^2 + q^2 + 8)$.

For a fixed $k \in \mathbb{N}^*$, we determine the solutions in $\mathbb{N}^* \times \mathbb{N}^*$ of the equation $p^2 + q^2 + 8 = kpq$ with $p, q < 2023$.

Let's assume that $(p_0, q_0)$ is a solution for which the sum $p+q$ is minimal and $p_0 \ge q_0$. If $p_0 = q_0$, since $p_0^2 \mid 8$, we have $p_0 = q_0 = 1$ or $p_0 = q_0 = 2$ (cases we will analyze later). Now, let's assume that for $p_0 \ge 3$, we have $p_0 > q_0$.

If $p_0 \ge 5$ and $p'$ is the second solution of the equation $x^2 - (q_0k)x + q_0^2 + 8 = 0$, then we have $p' = \frac{q_0^2+8}{p_0} \le \frac{p_0^2-2p_0+9}{p_0} < p_0$. Since $p' + q_0 < p_0 + q_0$, it follows that $p_0 \in \{3, 4\}$.

If $p_0 = 3$, then $q_0 \mid 17$, so $q_0 = 1$ and $k = 6$. Using Vieta jumping, we obtain the solutions given by the sequence $p_0 = 3, q_0 = 1, q_{n+1} = p_n, p_{n+1} = 6p_n - q_n$. Thus, we obtain the solutions $(3, 1)$, $(17, 3)$, $(99, 17)$, $(577, 99)$, while the remaining solutions have $p \ge 2023$. The only solution remaining is $(17, 3)$.

If $p_0 = 4$, then $q_0 \mid 24$ and $q_0 \le 3$, so $q_0$ is $2$ or $1$, but both cases are impossible.

Now, let's consider the cases $p_0 = 2$ and $p_0 = 1$.

If $p_0 = 2$, then $q_0 \mid 12$ and $q_0 \le 2$, so $q_0 = 2$ (the case $q_0 = 1$ is not possible) and $k = 4$. Using Vieta jumping, all solutions have both components even. The acceptable solution is $(2, 2)$.