Find all integer solutions $(p, q, r)$ of the equation $r + p ^ 4 = q ^ 4$ with the following conditions: 
$\bullet$ $r$ is a positive integer with exactly $8$ positive divisors. 
$\bullet$  $p$ and $q$ are prime numbers.
To find all integer solutions \((p, q, r)\) of the equation \(r + p^4 = q^4\) with the given conditions, we need to analyze the properties of \(r\), \(p\), and \(q\). 

1. **Determine the form of \(r\)**:
   Since \(r\) has exactly 8 positive divisors, it can be expressed in one of the following forms:
   - \(r = a^7\)
   - \(r = a^3b\)
   - \(r = abc\)
   where \(a, b, c\) are prime numbers.

2. **Factorize the equation**:
   The given equation can be rewritten as:
   \[
   r = q^4 - p^4
   \]
   Using the difference of squares, we can factorize \(q^4 - p^4\) as:
   \[
   q^4 - p^4 = (q^2 + p^2)(q^2 - p^2) = (q^2 + p^2)(q + p)(q - p)
   \]
   Therefore, we have:
   \[
   r = (q - p)(q + p)(q^2 + p^2)
   \]

3. **Analyze the gcd conditions**:
   Since \(p\) and \(q\) are prime numbers, we have \(\gcd(p, q) = 1\). We need to consider the gcd of the factors:
   - \(\gcd(q - p, q + p) \mid 2p \implies \gcd(q - p, q + p) \in \{1, 2\}\)
   - \(\gcd(q - p, q^2 + p^2) = \gcd(q - p, (q - p)(q + p) + 2p^2) \mid 2p^2 \implies \gcd(q - p, q^2 + p^2) \in \{1, 2\}\)
   - \(\gcd(q + p, q^2 + p^2) = \gcd(q + p, (q - p)(q + p) + 2p^2) \mid 2p^2 \implies \gcd(q + p, q^2 + p^2) \in \{1, 2\}\)

4. **Case 1: \(p = 2\)**:
   If \(p = 2\), then:
   \[
   r = (q - 2)(q + 2)(q^2 + 4)
   \]
   Since \(q\) is a prime number greater than 2, \(q\) must be odd, making \(q - 2\), \(q + 2\), and \(q^2 + 4\) all odd. Therefore, the gcd conditions are each 1. We must be in the case \(r = abc\).

   - If \(q - 2\) and \(q + 2\) are prime, then \(q = 5\) or \(q > 5\) and \(q \equiv 0 \pmod{3}\), which is impossible for a prime number \(q\) greater than 5. Therefore, \(q = 5\).

   Substituting \(p = 2\) and \(q = 5\):
   \[
   r = (5 - 2)(5 + 2)(5^2 + 4) = 3 \cdot 7 \cdot 29 = 609
   \]
   Thus, one solution is \((p, q, r) = (2, 5, 609)\).

5. **Case 2: \(p \geq 3\)**:
   If \(p \geq 3\), then \(q - p\), \(q + p\), and \(q^2 + p^2\) are all even, making the gcd conditions each 2. We must be in the case \(r = 2^3b\), which is clearly absurd since \((q - p)\), \((q + p)\), and \((q^2 + p^2)\) are all distinct and cannot all be powers of 2.

   Therefore, there are no more solutions.

The final answer is \( \boxed{ (2, 5, 609) } \).