Solution:

By checking, we find that for $n \leqslant 2$ the only solution is the pair $(1,2)$. Let us prove that for $n \geqslant 3$ there are no solutions.

The number $x$ must be odd, so $x^{2}+2 \equiv 3 \pmod{8}$. Now from $x\left(x^{2}+2\right) \equiv -1 \pmod{8}$ it follows that $x \equiv 5 \pmod{8}$. Moreover, since $3 \mid x\left(x^{2}+2\right)$ (if $3 \nmid x$ then $3 \mid x^{2}+2$), it must be that $2^{n} \equiv 1 \pmod{3}$, so $n$ is an even number.

Adding the number 2 to both sides of the equality gives
$$(x+1)\left(x^{2}-x+3\right)=2^{n}+2$$
Since $n$ is even, $2^{n}$ is a perfect square, so the number $-2$ is a quadratic residue modulo every odd prime divisor $p$ of the number $(x+1)\left(x^{2}-x+3\right)$. Therefore
$$1=\left(\frac{-2}{p}\right)=\left(\frac{-1}{p}\right)\left(\frac{2}{p}\right)=(-1)^{\frac{p-1}{2}}(-1)^{\frac{p^{2}-1}{8}}=(-1)^{\frac{(p-1)(p+5)}{8}}$$
from which it follows that $p$ is of the form $8k+1$ or $8k+3$. Being a product of such prime numbers, the number $x^{2}-x+3$ itself must also be of that form. However, since $x \equiv 5 \pmod{8}$, we have $x^{2}-x+3 \equiv 7 \pmod{8}$, which is a contradiction.

Therefore, the only solution of the given equation is $(x, n)=(1,2)$.