Determine all primes $p$ and all non-negative integers $m$ and $n$, such that $$1 + p^n = m^3. $$
We need to determine all primes \( p \) and all non-negative integers \( m \) and \( n \) such that 
\[ 1 + p^n = m^3. \]

1. **Initial Observations:**
   - Since \( m^3 \) is a cube, \( m \) must be at least 1.
   - If \( n = 0 \), then \( 1 + p^0 = m^3 \implies 2 = m^3 \). This is impossible since 2 is not a perfect cube. Therefore, \( n \geq 1 \).

2. **Case \( n = 1 \):**
   - If \( n = 1 \), then the equation becomes:
     \[ 1 + p = m^3. \]
   - Rearranging, we get:
     \[ p = m^3 - 1. \]
   - We need \( p \) to be a prime number. Let's factorize \( m^3 - 1 \):
     \[ m^3 - 1 = (m-1)(m^2 + m + 1). \]
   - For \( p \) to be prime, one of the factors must be 1 (since the product of two numbers is prime only if one of them is 1). Therefore, we have:
     \[ m - 1 = 1 \implies m = 2. \]
   - Substituting \( m = 2 \) back into the equation:
     \[ p = 2^3 - 1 = 7. \]
   - Thus, \( p = 7 \), \( n = 1 \), and \( m = 2 \) is a solution.

3. **Case \( n \geq 2 \):**
   - If \( n \geq 2 \), we can use Mihăilescu's Theorem (formerly known as Catalan's conjecture), which states that the only solution in the natural numbers of the equation \( x^a - y^b = 1 \) for \( a, b > 1 \) is \( x = 3, a = 2, y = 2, b = 3 \). 
   - Applying Mihăilescu's Theorem to our equation \( 1 + p^n = m^3 \), we see that there are no solutions for \( n \geq 2 \) because it would imply \( p^n = m^3 - 1 \), which contradicts the theorem unless \( p = 3 \), \( n = 2 \), \( m = 2 \), but this does not satisfy our equation.

4. **Conclusion:**
   - The only solution is when \( n = 1 \), \( m = 2 \), and \( p = 7 \).

The final answer is \( \boxed{ (p, n, m) = (7, 1, 2) } \).