18.6.1 ** Find infinitely many different sets of positive integers $x, y, z$, such that $x^{2}+1=y^{2}+z^{2}$, and the last several digits of $z^{2}$ are the same as those of $x^{2}+1$.
Let $x=y+2u, y=u^{3}$, then
$$
x^{2}+1-y^{2}=4u^{2}+4uy+1=4u^{4}+4u^{2}+1=\left(2u^{2}+1\right)^{2}.
$$

Therefore, $(x, y, z)=\left(u^{2}+2u, u^{3}, 2u^{2}+1\right)$ is the solution to this problem. Here, $u$ is any natural number ending with several zeros.