17 Find all positive integer triples $(x, y, z)$ that satisfy: $\left\{\begin{array}{l}x+y=z, \\ x^{2} y=z^{2}+1 .\end{array}\right.$
Solve the system of equations $\left\{\begin{array}{l}x+y=z, \\ x^{2} y=z^{2}+1\end{array}\right.$ to get $x^{2}(z-x)=z^{2}+1$, which simplifies to
$$
z^{2}-x^{2} z+x^{3}+1=0,
$$

treating $x$ as a parameter. When $x=1,2,3,4$, the discriminant of the quadratic equation (1) in $z$ is $\Delta z_{2} \Rightarrow z_{1}>\frac{x^{2}}{2} \Rightarrow z_{1} \geqslant \frac{x^{2}+1}{2}$, hence we have
$$
\begin{aligned}
z_{2} & =\frac{x^{3}+1}{z_{1}} \leqslant \frac{2\left(x^{3}+1\right)}{x^{2}+1}x^{3}+1,
\end{aligned}
$$

which implies $z_{2} \leqslant x+1$. Since $z_{1} z_{2}=\left(x^{2}-z_{2}\right) z_{2}=-z_{2}^{2}+x^{2} z_{2}$, when $x \geqslant 5$, $x+1<\frac{x^{2}}{2}$, so in the case $z_{2} \leqslant x+1$, the function $-z_{2}^{2}+x^{2} z_{2}$ is strictly increasing, thus
$$
\begin{aligned}
z_{1} z_{2}=\left(x^{2}-z_{2}\right) z_{2} & \leqslant\left(x^{2}-x-1\right)(x+1)=x^{3}-2 x-1 \\
& <x^{3}+1,
\end{aligned}
$$

which is a contradiction. Therefore, $z_{1} \leqslant(x-1)^{2}+2$, meaning $z_{1}$ can only take values $(x-1)^{2}, (x-1)^{2}+1, (x-1)^{2}+2$. Consequently, $z_{2}=x^{2}-z_{1}$ can only take values $2 x-1, 2 x-2, 2 x-3$. Substituting these into (1), we find that only when $z_{2}=2 x-3$, we get the integer $x=5$, thus $z_{2}=7, z_{1}=x^{2}-z_{2}=18 ; y_{1}=z_{1}-x=2, y_{2}=$ $z_{2}-x=13$, leading to the solutions: $(x, y, z)=(5,2,7)$ and $(5,3,18)$.