Solve the following equation in the set of natural numbers:

$$
\left(x^{2}+1\right)^{y}-\left(x^{2}-1\right)^{y}=2 x^{y}
$$
$$
\left(x^{2}+1\right)^{y}-\left(x^{2}-1\right)^{y}=2 x^{y}
$$

Divide both sides of the equation by the positive $\left(x^{2}+1\right)^{y}$, and regroup the terms:

$$
\left(\frac{x^{2}-1}{x^{2}+1}\right)^{y}+2\left(\frac{x}{x^{2}+1}\right)^{y}=1
$$

If $x \geq 1$, the bases on the left side are non-negative and less than 1, so the value on the left side is a monotonically decreasing function of $y$ for a fixed $x$. If $y=2$, the value on the left side is $\left(x^{4}+1\right) /\left(x^{4}+2 x^{2}+1\right)$, which is always less than 1, so only $y=1$ can be considered. In this case, the solution to (2) is $x=1$, so the only solution is: $x=y=1$.

Note. If $x=0$, and $y>0$, the left side of (2) is $(-1)^{y}$, which equals 1 for any even $y$. If $y=0$ and $x>0$, the value on the left side of (2) cannot be 1. If $x=y=0$, the second term is undefined. Therefore, if we include zero among the natural numbers, the set of solutions is expanded to $x=0$, $y$ any even number.