Find all pairs $(m,p)$ of natural numbers , such that $p$ is a prime and \[2^mp^2+27\] is the third power of a natural numbers
We need to find all pairs \((m, p)\) of natural numbers such that \(p\) is a prime and \(2^m p^2 + 27\) is the third power of a natural number. Let's denote this third power by \(x^3\), where \(x\) is a natural number. Thus, we have the equation:
\[ 2^m p^2 + 27 = x^3 \]

1. **Initial Solution Check:**
   Let's check if \(m = 1\) and \(p = 7\) works:
   \[ 2^1 \cdot 7^2 + 27 = 2 \cdot 49 + 27 = 98 + 27 = 125 = 5^3 \]
   This is indeed a solution.

2. **Case \(p = 2\):**
   If \(p = 2\), then the equation becomes:
   \[ 2^m \cdot 2^2 + 27 = x^3 \]
   \[ 2^{m+2} + 27 = x^3 \]
   Let \(x - 3 = 2^a\) and \(x^2 + 9 + 3x = 2^b\). Squaring the first equation and subtracting from the second, we get:
   \[ 9x = 2^b - 4^a \]
   Working modulo 2, if \(a \geq 1\) and \(b \geq 1\), then \(x\) is even, which contradicts \(x = 3 + 2^a\). If \(a = 0\), then \(x = 4\) and:
   \[ 16 + 9 + 12 = 37 \]
   which is not a power of 2. Thus, no solutions for \(p = 2\).

3. **Case 1: \((x - 3) = 2^a\) and \((x^2 + 9 + 3x) = 2^b p^2\):**
   Substituting \(x = 3 + 2^a\) into the second equation:
   \[ (3 + 2^a)^2 + 9 + 3(3 + 2^a) = 2^b p^2 \]
   \[ 9 + 6 \cdot 2^a + 4^a + 9 + 9 + 3 \cdot 2^a = 2^b p^2 \]
   \[ 27 + 9 \cdot 2^a + 4^a = 2^b p^2 \]
   If \(a \geq 1\) and \(b \geq 1\), working modulo 2, there are no solutions. If \(a = 0\) and \(b \geq 1\):
   \[ 37 = 2^b p^2 \]
   which has no solutions. If \(b = 0\) and \(a = 1\):
   \[ p = 7 \]
   Hence, one solution is \((m, p) = (1, 7)\). If \(b = 0\) and \(a \geq 1\), there are no solutions when working modulo 4.

4. **Case 2: \((x - 3) = p \cdot 2^a\) and \((x^2 + 9 + 3x) = p \cdot 2^b\):**
   Substituting into the second equation:
   \[ (3 + p \cdot 2^a)^2 + 9 + 3(3 + p \cdot 2^a) = p \cdot 2^b \]
   \[ 9 + 6p \cdot 2^a + p^2 \cdot 4^a + 9 + 9 + 3p \cdot 2^a = p \cdot 2^b \]
   \[ 27 + 9p \cdot 2^a + p^2 \cdot 4^a = p \cdot 2^b \]
   Working modulo \(p\), we see that \(p = 3\). Plugging \(p = 3\) and taking modulo 3, there are no solutions.

5. **Case 3: \((x - 3) = p^2 \cdot 2^a\) and \((x^2 + 9 + 3x) = 2^b\):**
   Substituting into the second equation:
   \[ (3 + p^2 \cdot 2^a)^2 + 9 + 3(3 + p^2 \cdot 2^a) = 2^b \]
   \[ 9 + 6p^2 \cdot 2^a + p^4 \cdot 4^a + 9 + 9 + 3p^2 \cdot 2^a = 2^b \]
   \[ 27 + 9p^2 \cdot 2^a + p^4 \cdot 4^a = 2^b \]
   Working modulo 2, we see that \(a = 0\) or \(b = 0\). But \(b \geq 4\) in this case, so \(a = 0\). Then working modulo 2 tells us that there are no solutions.

Hence, the only solution is \((m, p) = (1, 7)\).

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