4. Find all positive integer triples $(p, q, r)$ such that $p$ and $q$ are both primes, and $\frac{r^{2}-5 q^{2}}{p^{2}-1}=2$.

---

The translation maintains the original format and line breaks as requested.
4. $(p, q, r)=(3,2,6)$.

According to the problem, we have $r^{2}=2 p^{2}+5 q^{2}-2$, where $p$ and $q$ are both prime numbers.
If $p$ and $q$ are both not equal to 2, then
$$
\begin{array}{l}
p^{2} \equiv q^{2} \equiv 1(\bmod 8) \\
\Rightarrow 2 p^{2}+5 q^{2}-2 \equiv 5(\bmod 8),
\end{array}
$$

For any integer $r$, we have $r^{2} \equiv 0,1,4(\bmod 8)$,
thus, there does not exist a triplet $(p, q, r)$ that satisfies the condition in this case.
Similarly, if $p$ and $q$ are both not equal to 3, then
$$
\begin{array}{l}
p^{2} \equiv q^{2} \equiv 1(\bmod 3) \\
\Rightarrow 2 p^{2}+5 q^{2}-2 \equiv 2(\bmod 3),
\end{array}
$$

For any integer $r$, we have $r^{2} \equiv 0,1(\bmod 3)$, thus, there does not exist a triplet $(p, q, r)$ that satisfies the condition in this case either.

Therefore, we only need to substitute $(p, q)=(2,3),(3,2)$, $(2,2),(3,3)$ into the original equation for verification, and we find that the only triplet that meets the requirement is $(p, q, r)=(3,2,6)$.