Example 1 Find all triples of integers $(x, y, z)$ such that
$$
x^{3}+y^{3}+z^{3}-3 x y z=2003^{[1]} \text {. }
$$
$(2003$, Nordic Mathematical Contest)
Notice,
$$
\begin{aligned}
(x+y & +z)\left[(x-y)^{2}+(y-z)^{2}+(z-x)^{2}\right] \\
& =2 \times 2003,
\end{aligned}
$$

and $(x-y)^{2}+(y-z)^{2}+(z-x)^{2} \equiv 0(\bmod 2)$.
Thus, we have $\left\{\begin{array}{l}x+y+z=1, \\ (x-y)^{2}+(y-z)^{2}+(z-x)^{2}=4006 ;\end{array}\right.$

or $\left\{\begin{array}{l}x+y+z=2003, \\ (x-y)^{2}+(y-z)^{2}+(z-x)^{2}=2 \text {. }\end{array}\right.$
From equation set (1), we get
$$
\begin{array}{l}
(x-y)^{2}+(x+2 y-1)^{2}+(2 x+y-1)^{2} \\
=4006,
\end{array}
$$

which is $6 x^{2}+6 y^{2}+6 x y-6 x-6 y+2=4006$.
But $4006 \equiv 4(\bmod 6)$, which is a contradiction.
Now, let's discuss equation set (2).
Since $|x-y|, |y-z|, |z-x|$ have two 1s and one 0, without loss of generality, assume $x \geqslant y \geqslant z$.
When $x-1=y=z$, $3 y+1=2003$, no solution;
When $x=y=z+1$,
$$
3 x-1=2003 \Rightarrow x=668 \text {. }
$$

Therefore, the ternary tuples that satisfy the conditions are
$$
\begin{array}{l}
(x, y, z) \\
=(668,668,667),(668,667,668), \\
\quad(667,668,668) .
\end{array}
$$