Find all triples $ \left(p,x,y\right)$ such that $ p^x\equal{}y^4\plus{}4$, where $ p$ is a prime and $ x$ and $ y$ are natural numbers.
1. Start with the given equation:
   \[
   p^x = y^4 + 4
   \]
2. Factorize the right-hand side:
   \[
   y^4 + 4 = (y^2 - 2y + 2)(y^2 + 2y + 2)
   \]
3. Since \( p \) is a prime, \( p^x \) must be a power of \( p \). Therefore, both factors on the right-hand side must be powers of \( p \):
   \[
   p^a = y^2 + 2y + 2 \quad \text{and} \quad p^b = y^2 - 2y + 2 \quad \text{where} \quad a > b , \quad a + b = x \in \mathbb{N}
   \]
4. Subtract the second equation from the first:
   \[
   p^a - p^b = (y^2 + 2y + 2) - (y^2 - 2y + 2) = 4y
   \]
   Thus,
   \[
   y = \frac{p^a - p^b}{4}
   \]
5. Add the two equations:
   \[
   p^a + p^b = (y^2 + 2y + 2) + (y^2 - 2y + 2) = 2(y^2 + 2)
   \]
   Substituting \( y = \frac{p^a - p^b}{4} \):
   \[
   p^a + p^b = 2\left( \left(\frac{p^a - p^b}{4} \right)^2 + 2\right)
   \]
6. Simplify the equation:
   \[
   p^a + p^b = 2\left( \frac{(p^a - p^b)^2}{16} + 2 \right)
   \]
   \[
   8(p^a + p^b) = (p^a - p^b)^2 + 32
   \]

**Case 1: \( b > 1 \)**

7. If \( b > 1 \), then \( p^b \) is divisible by \( p \):
   \[
   32 \equiv 0 \pmod{p} \implies p \mid 32 \implies p = 2
   \]
8. If \( p = 2 \), then:
   \[
   2^x = y^4 + 4
   \]
   Since \( y \) must be even, let \( y = 2m \):
   \[
   2^x = 16m^4 + 4 \implies 4(4m^4 + 1) = 2^x \implies x = 2 + 2k \quad \text{(for some integer \( k \))}
   \]
   However, this does not yield a valid solution for \( x \) in natural numbers. Thus, there is no solution in this case.

**Case 2: \( b = 0 \)**

9. If \( b = 0 \), then:
   \[
   (p^a - 1)^2 + 32 = 8(p^a + 1)
   \]
10. Simplify the equation:
    \[
    p^{2a} - 2p^a + 33 = 8p^a + 8
    \]
    \[
    p^{2a} - 10p^a + 25 = 0
    \]
11. Factorize the quadratic equation:
    \[
    (p^a - 5)^2 = 0 \implies p^a = 5 \implies p = 5, \quad a = 1
    \]
12. Substitute \( p = 5 \) and \( a = 1 \) back into the equation for \( y \):
    \[
    5^1 = y^2 + 2y + 2 \implies 5 = y^2 + 2y + 2
    \]
    \[
    y^2 + 2y - 3 = 0
    \]
    \[
    (y + 3)(y - 1) = 0 \implies y = 1 \quad (\text{since } y \in \mathbb{N})
    \]
13. Therefore, the solution is:
    \[
    (p, x, y) = (5, 1, 1)
    \]

The final answer is \(\boxed{(5, 1, 1)}\).