Proof:
We prove the statement by induction on $z$.
When $z=1$, $xy=2$, so $(x, y)=(1,2)$ or $(2,1)$. Thus, $(x, y, z)=$ $(1,2,1)$ or $(2,1,1)$. By setting
$$(a, b, c, d)=(1,0,1,1),(1,1,0,1)$$

we see that the proposition holds for $z=1$.
Now suppose $(x_0, y_0, z_0)$ is a positive integer solution to (17), and that (17) holds for $z_1$.

We need to show that there exist $a, b, c, d \in \mathbf{Z}$ such that
$$x_0=a^2+b^2, y_0=c^2+d^2, z_0=ac+bd$$

with $|ad-bc|=1$.
Indeed, for this positive integer solution $(x_0, y_0, z_0)$, where $z_0 > 1$, without loss of generality, assume $x_0 \leqslant y_0$. Consider the tuple
$$(x_1, y_1, z_1) = (x_0, x_0 + y_0 - 2z_0, z_0 - x_0).$$

Here, $x_1 = x_0 > 0$, and $y_1 = x_0 + y_0 - 2z_0 \geqslant 2\sqrt{x_0 y_0} - 2z_0 = 2\sqrt{z_0^2 + 1} - 2z_0 > 0$. For $z_1$, we have $x_0^2 \leqslant x_0 y_0 = z_0^2 + 1$, so $x_0 \leqslant \sqrt{z_0^2 + 1}$. Therefore, $x_0 \leqslant z_0$. If $x_0 = z_0$, then $z_0 \mid (z_0^2 + 1)$ implies $z_0 \mid 1$, leading to $z_0 = 1$, a contradiction. Hence, $z_1 = z_0 - x_0 > 0$, meaning $x_1, y_1, z_1$ are all positive integers.

Furthermore,
$$\begin{aligned} x_1 y_1 & = x_0 (x_0 + y_0 - 2z_0) = x_0^2 + x_0 y_0 - 2z_0 x_0 \\ & = x_0^2 + z_0^2 + 1 - 2x_0 z_0 \\ & = (z_0 - x_0)^2 + 1 = z_1^2 + 1 \end{aligned}$$

By the induction hypothesis, there exist $p, q, m, n \in \mathbf{Z}$ such that
$$(x_1, y_1, z_1) = (p^2 + q^2, m^2 + n^2, pm + qn)$$

with $|pn - qm| = 1$.
Now let $(a, b, c, d) = (p, q, p + m, q + n)$, then
$$\begin{aligned} x_0 & = x_1 = p^2 + q^2 = a^2 + b^2 \\ y_0 & = y_1 + x_1 + 2z_1 = p^2 + q^2 + m^2 + n^2 + 2(pm + qn) \\ & = (p + m)^2 + (q + n)^2 = c^2 + d^2 \\ z_0 & = x_1 + z_1 = p^2 + q^2 + pm + qn = p(p + m) + q(q + n) \\ & = ac + bd \end{aligned}$$

Finally,
$$|ad - bc| = |p(q + n) - q(p + m)| = |pn - qm| = 1$$

Thus, the proposition is proved.