We are tasked with finding all quadruples of positive integers $(p, q, a, b)$, where $p$ and $q$ are prime numbers, $a > 1$, and they satisfy the equation:

$$p^a = 1 + 5q^b.$$

Given the integers involved, we will first check small values for $p$, and test if they can satisfy the equation.

### Case 1: $p = 2$

If $p = 2$, the equation becomes:

$$2^a = 1 + 5q^b.$$

Rearranging gives:

$$2^a - 1 = 5q^b.$$

Check small values of $a$ starting from $a = 2$:

- **For $a = 2$:**
$$2^2 - 1 = 3 \neq 5q^b.$$

- **For $a = 3$:**
$$2^3 - 1 = 7 \neq 5q^b.$$

- **For $a = 4$:**
$$2^4 - 1 = 15 = 5 \cdot 3^1.$$

Here we find a solution: $(p, q, a, b) = (2, 3, 4, 1)$.

### Case 2: $p = 3$

If $p = 3$, the equation becomes:

$$3^a = 1 + 5q^b.$$

Rearranging gives:

$$3^a - 1 = 5q^b.$$

Check small values of $a$:

- **For $a = 2$:**
$$3^2 - 1 = 8 \neq 5q^b.$$

- **For $a = 3$:**
$$3^3 - 1 = 26 \neq 5q^b.$$

- **For $a = 4$:**
$$3^4 - 1 = 80 = 5 \cdot 2^4.$$

Here we find another solution: $(p, q, a, b) = (3, 2, 4, 4)$.

### Verification for Other Possible Primes

Beyond $p = 3$, the exponential growth of $p^a$ becomes too large rapidly compared to the relatively small possibilities of $5q^b + 1$, thus unlikely to satisfy the equation. Therefore, we focus on small primes and small values for $a$.

### Conclusion

The solutions $(p, q, a, b)$ that satisfy the equation are:

$$\boxed{(2, 3, 4, 1) \text{ and } (3, 2, 4, 4)}.$$

These are the only solutions that meet the conditions of the given problem.