Find all triples $(p,n,k)$ of positive integers, where $p$ is a Fermat's Prime, satisfying \[p^n + n = (n+1)^k\].

[i]Observation: a Fermat's Prime is a prime number of the form $2^{\alpha} + 1$, for $\alpha$ positive integer.[/i]
1. **Identify Fermat's Primes**: Fermat's primes are of the form \( p = 2^{2^r} + 1 \) for non-negative integers \( r \). The known Fermat primes are \( 3, 5, 17, 257, \) and \( 65537 \).

2. **Case \( n = 1 \)**:
   \[
   p^1 + 1 = (1+1)^k \implies p + 1 = 2^k
   \]
   Since \( p \) is a Fermat prime, we test the known Fermat primes:
   - For \( p = 3 \):
     \[
     3 + 1 = 4 = 2^2 \implies k = 2
     \]
     This gives the solution \( (3, 1, 2) \).

3. **Case \( n > 1 \)**:
   - Let \( q \) be the smallest prime dividing \( n \). Taking the equation modulo \( q \):
     \[
     p^n + n \equiv (n+1)^k \pmod{q}
     \]
     Since \( q \mid n \), we have:
     \[
     p^n \equiv -n \pmod{q} \implies p^n \equiv 0 \pmod{q} \implies p \equiv 1 \pmod{q}
     \]
     This implies \( q = 2 \) because \( p \) is a Fermat prime and \( p-1 \) is a power of 2.

4. **Rewrite the equation**:
   \[
   p^n - 1 = (n+1)^k - n - 1
   \]
   Factorize the right-hand side:
   \[
   p^n - 1 = (n+1)((n+1)^{k-1} - 1)
   \]

5. **Valuation at 2**:
   - If \( k-1 \) is odd:
     \[
     v_2(p^n - 1) = v_2((n+1)^2 - 1) + v_2(k-1) - 1
     \]
     Since \( v_2(p^2 - 1) \geq 3 \), this leads to a contradiction. Thus, \( k-1 \) must be even.

6. **Valuation at 2 continued**:
   \[
   v_2(p^n - 1) = v_2((n+1)^2 - 1) + v_2(k-1) - 1
   \]
   \[
   v_2(k-1) = v_2(p^2 - 1) - v_2(n+2)
   \]
   Since \( 2^\alpha = p-1 \nmid k-1 \), we have \( p > 3 \), so \( p \equiv 1 \pmod{4} \).

7. **Modulo 5 analysis**:
   - For \( p > 5 \):
     \[
     p \equiv 2 \pmod{5}
     \]
     \[
     4 + n \equiv (n+1)^k \pmod{5}
     \]
     This implies \( k \equiv 3 \pmod{4} \), leading to no solutions.

8. **Case \( p = 5 \)**:
   - \( v_2(k-1) + v_2(n+2) = 3 \):
     \[
     v_2(k-1) = 1 \quad \text{and} \quad v_2(n+2) = 2 \implies n \equiv 2 \pmod{8}
     \]
     \[
     n \equiv (n+1)^3 \pmod{5}
     \]
     This has the solution \( n \equiv 2 \pmod{5} \).

9. **Check \( n = 2 \)**:
   \[
   5^2 + 2 = 27 = 3^3
   \]
   This gives the solution \( (5, 2, 3) \).

10. **For \( n > 2 \)**:
    - \( n \) must have a smallest odd prime divisor \( q \):
      \[
      q \mid 5^{(n, q-1)} - 1 = 24 \implies q = 3
      \]
      \[
      1 + v_3(n/2) = v_3(n) + v_3(k-1) \implies k \equiv 7 \pmod{12}
      \]
      \[
      n \equiv 2 \pmod{4} \implies -1 + n \equiv (n+1)^7 \pmod{13}
      \]
      This leads to a contradiction.

In conclusion, the only solutions are \( (3, 1, 2) \) and \( (5, 2, 3) \).

The final answer is \( \boxed{ (3, 1, 2) } \) and \( (5, 2, 3) \).