We are tasked with finding all prime numbers $p, q, r$ that satisfy the equation:

$$\frac{p}{q} - \frac{4}{r+1} = 1.$$

First, we rearrange the equation to find a common denominator:

$$\frac{p}{q} - \frac{4}{r+1} = 1 \implies \frac{p(r+1) - 4q}{q(r+1)} = 1.$$

This simplifies to:

$$p(r+1) - 4q = q(r+1).$$

Rearranging gives:

$$p(r+1) - q(r+1) = 4q \implies (p-q)(r+1) = 4q.$$

Since $p, q,$ and $r$ are prime numbers, consider the divisors of $4q$. Given the factors are affected by the nature of primes, let's analyze some simpler cases where $q$ and $r+1$ are constrained to familiar factors of small primes:

1. **Case $q = 2$:**

$$(p-2)(r+1) = 8.$$

- $r+1 = 2$: Then $r = 1$, not prime.
- $r+1 = 4$: Then $r = 3$, and hence $(p-2) = 2$ implies $p = 4$, not prime.
- $r+1 = 8$: Then $r = 7$, and hence $(p-2) = 1$ implies $p = 3$.

If $(p, q, r) = (3, 2, 7)$, this is verified:
$$\frac{3}{2} - \frac{4}{8} = \frac{3}{2} - \frac{1}{2} = 1.$$

2. **Case $q = 3$:**

$$(p-3)(r+1) = 12.$$

- $r+1 = 3$: Then $r = 2$, and hence $(p-3) = 4$ implies $p = 7$.
- $r+1 = 4$: Then $r = 3$, and hence $(p-3) = 3$ implies $p = 6$, not prime.
- $r+1 = 6$: Then $r = 5$, and hence $(p-3) = 2$ implies $p = 5$.

So we get two solutions $(p, q, r) = (7, 3, 2)$ and $(5, 3, 5)$.

Verifications:
$$\frac{7}{3} - \frac{4}{3} = \frac{7}{3} - \frac{4}{3} = 1.$$
$$\frac{5}{3} - \frac{4}{6} = \frac{5}{3} - \frac{2}{3} = 1.$$

Thus, all possible valid solutions for primes $(p, q, r)$ are:

$$\boxed{(7, 3, 2), (5, 3, 5), (3, 2, 7)}.$$