Problem 1271

To prove, for the subsequent argument, we first derive some simple conclusions from equation (1).
Clearly, $n>1$. Moreover, $x$ must be odd; otherwise, taking (1) modulo 4 leads to a contradiction. Furthermore, $n$ is also odd, because if $2 \mid n$, then $x^{n}$ is the square of an odd number, making the right side of (1) $\equiv 1+2=3(\bmod 4)$, while the left side $\equiv 1(\bmod 4)$, which is impossible. Hence, $2 \nmid n$.

Let $x+1=2^{\alpha} x_{1}$, where $x_{1}$ is odd and $\alpha>0$ (since $x$ is odd). Rewrite equation (1) as
$$(x+2)^{2 m}-1=x^{n}+1$$
The left side of (2) has a factor $(x+2)^{2}-1=\left(2^{\alpha} x_{1}+1\right)^{2}-1=2^{a+1}\left(2^{\sigma-1} x_{1}^{2}+x_{1}\right)$, so $2^{a+1}$ divides the left side of (2). On the other hand, since $n-1>0$ is even, using the binomial theorem, we easily get
$$x^{n}+1=x\left(2^{a} x_{1}-1\right)^{n-1}+1 \equiv x \cdot 1+1=2^{a} x_{1}\left(\bmod 2^{a+1}\right)$$

Since $2 \nmid x_{1}$, the right side of (2) $x^{n}+1 \not \equiv 0\left(\bmod 2^{a+1}\right)$, leading to a contradiction!