One, (Full marks 20 points) Given that $x$, $y$, $z$ are positive integers, and satisfy $x^{3}-y^{3}-z^{3}=3 x y z$, $x^{2}=2(y+z)$. Find the value of $x y+y z+z x$.

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The translation is provided as requested, maintaining the original format and line breaks.
$$
\begin{aligned}
- & \because x^{3}-y^{3}-z^{3}-3 x y z \\
= & x^{3}-(y+z)^{3}-3 x y z+3 y^{2} z+3 y z^{2} \\
= & (x-y-z)\left(x^{2}+x y+x z+y^{2}+2 y z+z^{2}\right) \\
& -3 y z(x-y-z) \\
= & (x-y-z)\left(x^{2}+y^{2}+z^{2}-x y-y z+x z\right),
\end{aligned}
$$
X. $x^{3}-y^{3}-z^{3}=3 x y z$,
$$
\begin{array}{l}
\therefore(x-y-z)\left(x^{2}+y^{2}+z^{2}+x y-y z+x z\right) \\
\quad=0 .
\end{array}
$$
Since $x, y, z$ are all natural numbers,
$$
\begin{array}{l}
\text { and } x^{2}+y^{2}+z^{2}+x y-y z+x z \\
=\frac{1}{2}\left[(x+y)^{2}+(y-z)^{2}+(z+x)^{2} \neq 0,\right. \\
\therefore x-y-z=0,
\end{array}
$$

i.e., $x=y+z$.
Substituting (1) into $x^{2}=2(y+z)$, we get $x^{2}=2 x$.
$$
\therefore x=2, y+z=2 \text {. }
$$

Given $y \geqslant 1$ and $z \geqslant 1$, we know $y=z=1$.
$$
\therefore x y+y z+z x=5 \text {. }
$$