Find all triples $(x,n,p)$ of positive integers $x$ and $n$ and primes $p$ for which the following holds $x^3 + 3x + 14 = 2 p^n$
We are given the equation \( x^3 + 3x + 14 = 2p^n \) and need to find all triples \((x, n, p)\) of positive integers \(x\) and \(n\) and primes \(p\) that satisfy this equation.

1. **Rewrite the equation:**
   \[
   x^3 + 3x + 14 = 2p^n
   \]
   This can be factored as:
   \[
   (x + 2)(x^2 - 2x + 7) = 2p^n
   \]

2. **Define \( f(x) \):**
   Let \( f(x) = x^2 - 2x + 7 \). We need to find the greatest common divisor (GCD) of \( x + 2 \) and \( f(x) \):
   \[
   d = \gcd(x + 2, f(x))
   \]

3. **Evaluate \( f(-2) \):**
   \[
   f(-2) = (-2)^2 - 2(-2) + 7 = 4 + 4 + 7 = 15
   \]
   Since \( d \mid f(-2) \), we have:
   \[
   d \mid 15
   \]
   Therefore, \( d \) can be \( 1, 3, 5, \) or \( 15 \). However, since \( f(x) > x + 2 > 2 \), \( d \) must be a prime factor of 15, i.e., \( d = 3 \) or \( d = 5 \).

4. **Consider the parity of \( x \):**
   - If \( x \) is even, then \( x = 2k \) for some integer \( k \). Substituting \( x = 2k \) into the equation:
     \[
     (2k + 2)(4k^2 - 4k + 7) = 2p^n
     \]
     Since \( 2k + 2 \) is even, \( 4k^2 - 4k + 7 \) must be odd. This implies \( p^n \equiv 7 \pmod{8} \), which is impossible because \( 3^k \equiv 1, 3 \pmod{8} \) and \( 5^m \equiv 1, 5 \pmod{8} \) for all non-negative integers \( k \) and \( m \). Hence, \( x \) cannot be even.

5. **Conclude \( x \) is odd:**
   Since \( x \) is odd, we have:
   \[
   (x + 2, f(x)) = (p, 2p^n)
   \]
   This implies:
   \[
   x + 2 = p \quad \text{and} \quad f(x) = 2p^n
   \]

6. **Substitute \( x = p - 2 \):**
   \[
   f(p - 2) = (p - 2)^2 - 2(p - 2) + 7 = p^2 - 4p + 4 - 2p + 4 + 7 = p^2 - 6p + 11
   \]
   Therefore:
   \[
   2p^n = p^2 - 6p + 11
   \]

7. **Solve for \( p \) and \( n \):**
   - For \( p = 3 \):
     \[
     2 \cdot 3^n = 3^2 - 6 \cdot 3 + 11 = 9 - 18 + 11 = 2
     \]
     \[
     2 \cdot 3^n = 2 \implies 3^n = 1 \implies n = 0 \quad (\text{not valid since } n \text{ must be positive})
     \]
   - For \( p = 5 \):
     \[
     2 \cdot 5^n = 5^2 - 6 \cdot 5 + 11 = 25 - 30 + 11 = 6
     \]
     \[
     2 \cdot 5^n = 6 \implies 5^n = 3 \quad (\text{not valid since } 5^n \text{ must be a power of 5})
     \]

8. **Check specific values:**
   - For \( x = 1 \):
     \[
     1^3 + 3 \cdot 1 + 14 = 1 + 3 + 14 = 18 = 2 \cdot 3^2 \implies (x, n, p) = (1, 2, 3)
     \]
   - For \( x = 3 \):
     \[
     3^3 + 3 \cdot 3 + 14 = 27 + 9 + 14 = 50 = 2 \cdot 5^2 \implies (x, n, p) = (3, 2, 5)
     \]

Thus, the only solutions are \((x, n, p) = (1, 2, 3)\) and \((x, n, p) = (3, 2, 5)\).

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