Example 2 Find all integer triples $(x, y, z)$ such that $x^{3}+y^{3}+z^{3}-3 x y z=2003$.

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The left side of the equation is a familiar model, that is,
$$
\begin{array}{l}
x^{3}+y^{3}+z^{3}-3 x y z \\
=(x+y+z)\left(x^{2}+y^{2}+z^{2}-x y-y z-x z\right) .
\end{array}
$$

Notice that,
$$
\begin{array}{l}
(x-y)^{2}+(y-z)^{2}+(x-z)^{2} \\
=2\left(x^{2}+y^{2}+z^{2}-x y-y z-x z\right) .
\end{array}
$$

And $(x-y)^{2}+(y-z)^{2}+(x-z)^{2}$ is always an even number, i.e., its parity is an invariant, so the solution approach is obvious.
Transform the equation into
$$
\begin{array}{l}
(x+y+z)\left[(x-y)^{2}+(y-z)^{2}+(x-z)^{2}\right] \\
=2 \times 2003 . \\
\text { Then }\left\{\begin{array}{l}
x+y+z=1, \\
(x-y)^{2}+(y-z)^{2}+(x-z)^{2}=4006 ;
\end{array}\right.
\end{array}
$$

or $\left\{\begin{array}{l}x+y+z=2003, \\ (x-y)^{2}+(y-z)^{2}+(x-z)^{2}=2 .\end{array}\right.$
From equation set (1) we get
$$
\begin{array}{l}
(x-y)^{2}+(x+2 y-1)^{2}+(2 x+y-1)^{2}=4006 \\
\Rightarrow 6 x^{2}+6 y^{2}+6 x y-6 x-6 y+2=4006 \\
\Rightarrow 2=4(\bmod 6),
\end{array}
$$

which is a contradiction.
From equation set (2) we get $|x-y|, |y-z|, |x-z|$ have two values of 1 and one value of 0.
Therefore, the integer triplets that satisfy the conditions are
$$
\begin{aligned}
(x, y, z)= & (668,668,667),(668,667,668), \\
& (667,668,668) .
\end{aligned}
$$