To find all triples $(p, q, r)$ of positive integers, where $p, q$ are also primes, such that:

$$\frac{r^2 - 5q^2}{p^2 - 1} = 2,$$

we start by rearranging the equation:

$$r^2 - 5q^2 = 2(p^2 - 1).$$

This can be further rewritten as:

$$r^2 = 5q^2 + 2(p^2 - 1).$$

Since $p$ and $q$ are primes, we will consider small prime values for $p$ and $q$ and verify whether such values result in a perfect square for $r^2$.

Step-by-step Trial:

1. Let $p = 3$, since the smallest prime is too restrictive (as $p^2 - 1 = 0$ for $p=2$):

$$p^2 - 1 = 3^2 - 1 = 9 - 1 = 8.$$

Plug this back into the original equation:

$$r^2 = 5q^2 + 2 \times 8 = 5q^2 + 16.$$

2. For small prime $q = 2$:

$$5q^2 = 5 \times 2^2 = 20.$$

Substituting into the equation gives:

$$r^2 = 5 \times 4 + 16 = 20 + 16 = 36.$$

This is a perfect square, as $r = 6$.

Thus, a valid solution to the equation $\frac{r^2 - 5q^2}{p^2 - 1} = 2$ is $(p, q, r) = (3, 2, 6)$.

Therefore, the triples satisfying the given equation are:
$$\boxed{(3, 2, 6)}.$$

No other small prime values yield another valid integer solution for $r$ because further increases in $p$ rapidly increase $p^2 - 1$, which in turn would require significant increases in either $q$ or $r$, resulting in either non-integers or non-primes.