18.2.15 ** Find the positive integer triplet $(x, y, z)$, satisfying
$$
\left\{\begin{array}{l}
x+y=z, \\
x^{2} y=z^{2}+1 .
\end{array}\right.
$$
Parse 1: Given $y = z - x$, substituting into the second equation yields a quadratic equation in $z$:
$$
z^{2} - x^{2} z + x^{3} + 1 = 0.
$$

The discriminant $\Delta = x^{4} - 4(x^{3} + 1) = x^{4} - 4x^{3} - 4 > 0$, so (1) has two distinct roots $u > v$.
$u > \frac{x^{2}}{2}$, hence $u \geq \frac{x^{2} + 1}{2}, v = \frac{x^{3} + 1}{u} \leq \frac{2(x^{3} + 1)}{x^{2} + 1}$. 

\[
(2x - 2)(x^{2} - 2x + 2) = x^{3} + 1 + x^{3} - 2x^{2} + 3 > x^{3} + 1, \\
(2x - 3)(x^{2} - 2x + 3) = x^{3} + 1 + x^{3} - 7x^{2} + 12x - 10 \\
= x^{3} + 1 + (x - 5)(x^{2} - 2x + 2).
\]

Thus, the only solution is $x = 5, u = 18, v = 7$. The solutions to the problem are:
$$
\left\{\begin{array}{l}
x = 5 \\
y = 2 \\
z = 7
\end{array} \quad \left\{\begin{array}{l}
x = 5 \\
y = 13 \\
z = 18
\end{array}\right.\right.
$$

Parse 2: Substituting $z = x + y$ into the second equation gives:
$$
x^{2} y = (x + y)^{2} + 1.
$$
When $y = 1$, $x^{2} y = x^{2} \geq 4x$, so $x \geq 5$.
The quadratic function in $y$, $y^{2} + (2x - x^{2})y + x^{2} + 1$, achieves its maximum value at $y = 2$ or $y = \frac{x^{2} + 1}{2}$. Therefore, $y^{2} + (2x - x^{2})y + x^{2} + 1 \leq 4 + 2(2x - x^{2}) + x^{2} + 1 = 5 + 4x - x^{2} \leq 5 - x \leq 0$, or $y^{2} + (2x - x^{2})y + x^{2} + 1 = \frac{x^{2} + 1}{4}(x^{2} + 1 + 2(2x - x^{2}) + 1) \leq 0$. Equality holds only when $x = 5, y = 2$ or $\left(\frac{x^{2} + 1}{2} = 13\right)$. Therefore, the solutions to the problem are:
$$
\left\{\begin{array}{l}
x = 5 \\
y = 2, \\
z = 7, 
\end{array} \quad \left\{\begin{array}{l}
x = 5 \\
y = 13 \\
z = 18
\end{array}\right.\right.
$$