Clearly, this equation is insolvable when $n$ is even.

When $n=1$, there exists one solution $x=4$.

When $n$ is odd and greater than 1, the equation can be expanded to a polynomial in the following form: $$x^n + \cdots + 2^{n+1} = 0$$

Rational root theorem assets that any integral root must be a divisor of $2^{n+1}$. When $x > 0$, it is easy to show $x^n + (2+x)^n + (2-x)^n > 0$. Hence, any possible root must be in the form of $- 2^k$ where $k$ is a non-negative integer not exceeding $(n+1)$.

• When $k=0, 1, 2$, it can be verified that none of $x=-1, -2,  -4$ is a root.
• When $k\ge 3$, then setting $x=-2^k$ leads to $$\begin{array}{rl} &-2^{nk} + (2-2^k)^n + (2+2^k)^n = 0\\ \Leftrightarrow& -2^{n(k-1)} + (1-2^{k-1})^n + (1+2^{k-1})^n = 0\end{array}$$

Noting $k \ge 3$ and $n$ is a positive integer, the left side of the last equation is congruent to $2$ modulo $4$. However, the right side $0\equiv 0\pmod{4}$. Therefore, this equation is insolvable.

In conclusion, the given equation is only solvable when $n=\boxed{1}$.