Answer: $(x; n; p) = (1; 2; 3)$, $(x; n; p) = (3; 2; 5)$.

It is easy to see that $x^3 + 3x + 14 = (x+2)(x^2 - 2x + 7)$, so the initial equality can be rewritten as
$$(x+2)(x^2-2x+7)=2 \cdot p^n \quad (1)$$
It is evident that $x^2-2x+7 > x+2$ for all $x \in \mathbb{N}$. So for the case $x+2 = 2 \cdot p^k$, $x^2 - 2x + 7 = 2 \cdot p^{n-k}$ we have $n-k \ge k \ge 0$. It means that for both the cases $2(x^2 - 2x + 7) : (x+2)$. Therefore, the number $2(x^2 - 2x + 7)/(x+2)$ is integer, but since $2(x^2 - 2x + 7) = 2x(x+2) - 8(x+2) - 30$, the number $30/(x+2)$ must be integer. It possible only if $(x+2)$ is a divisor of $30$. From (1) it follows that the number $x+2$ has at most two prime divisors, one of them being $2$. So for the natural number $x$ the number $x+2$ can admit only the following values $3, 5, 6, 10$, i.e. $x$ can admit only the values $1, 3, 4, 8$.

For $x=1$ we have $(x+2)(x^2-2x+7)=3 \cdot 6=18=2 \cdot 3^2$, so $p=3$, $n=2$.

For $x=3$ we have $(x+2)(x^2-2x+7)=5 \cdot 10=2 \cdot 5^2$, so $p=5$, $n=2$.

For $x=4$ we have $(x+2)(x^2-2x+7)=6 \cdot 15=2 \cdot 3^2 \cdot 5$, i.e. this number cannot be $2 \cdot p^n$ for any natural $n$ and any prime $p$.

For $x=8$ we have $(x+2)(x^2-2x+7)=10 \cdot 55=2 \cdot 5^2 \cdot 11$, i.e. this number cannot be $2 \cdot p^n$ for any natural $n$ and any prime $p$.