Let $p$ be a prime number, determine all positive integers $(x, y, z)$ such that:
$x^p + y^p = p^z$
1. **Assume \( p \mid x \):**
   - Let \( p^a \parallel x \) and \( p^b \parallel y \), where \( a \leq b \).
   - We have:
     \[
     x = p^a x' \quad \text{and} \quad y = p^b y'
     \]
     where \( p \nmid x' \) and \( p \nmid y' \).
   - Substituting these into the original equation:
     \[
     (p^a x')^p + (p^b y')^p = p^z
     \]
     which simplifies to:
     \[
     p^{ap} x'^p + p^{bp} y'^p = p^z
     \]
     Dividing through by \( p^{ap} \):
     \[
     x'^p + p^{bp-ap} y'^p = p^{z-ap}
     \]
   - If \( z = ap \), then:
     \[
     x'^p + p^{bp-ap} y'^p = 1
     \]
     which is a contradiction since \( x'^p \geq 1 \) and \( p^{bp-ap} y'^p \geq 1 \), making the left-hand side at least 2.
   - If \( z > ap \), then:
     \[
     x'^p + p^{bp-ap} y'^p = p^{z-ap}
     \]
     If \( bp > ap \), then \( bp - ap > 0 \), and \( p \mid x' \), which is a contradiction since \( p \nmid x' \). Hence, \( b = a \), and we can assume \( p \nmid x \) and \( p \nmid y \).

2. **Assume \( p \nmid x \) and \( p \nmid y \):**
   - Rewrite the equation:
     \[
     x^p + y^p = p^z
     \]
   - Factorize the left-hand side:
     \[
     (x+y)(x^{p-1} - x^{p-2}y + \cdots + y^{p-1}) = p^z
     \]
   - Study the largest exponent of \( p \) dividing \( x+y \):
     - If \( x+y \equiv 0 \pmod{p^2} \), then \( x \equiv -y \pmod{p^2} \):
       \[
       x^{p-1} - x^{p-2}y + \cdots + y^{p-1} \equiv px^{p-1} \pmod{p^2}
       \]
       Since \( p \nmid x \), by Fermat's Little Theorem (FLT), \( x^{p-1} \equiv 1 \pmod{p} \):
       \[
       px^{p-1} \equiv p \pmod{p^2}
       \]
       Hence, \( x^{p-1} - x^{p-2}y + \cdots + y^{p-1} \) is divisible by \( p \) but not by \( p^2 \).

3. **Consider \( x+y = p \):**
   - Let \( y = p - x \):
     \[
     x^p + (p-x)^p = p^z
     \]
   - Using the binomial expansion:
     \[
     (p-x)^p = \sum_{i=0}^p \binom{p}{i} p^i (-x)^{p-i}
     \]
     For \( i \geq 3 \), \( \binom{p}{i} p^i (-x)^{p-i} \equiv 0 \pmod{p^3} \):
     \[
     x^p + (p-x)^p = x^p + \left( p^p - \binom{p}{1} p x + \binom{p}{2} p^2 x^{p-2} + \cdots \right)
     \]
     Simplifying:
     \[
     x^p + (p-x)^p \equiv x^p + p^2 x^{p-1} \pmod{p^3}
     \]
     Hence:
     \[
     x^p + (p-x)^p = p^2 (u p + x^{p-1})
     \]
     For some integer \( u \), implying \( z = 2 \).

Conclusion:
The only solution is when \( z = 2 \).

The final answer is \( \boxed{ z = 2 } \).