Example 2 Find all integer triples $(x, y, z)$ such that:
$$
x^{3}+y^{3}+z^{3}-3 x y z=2003 .
$$
(2003 17th Nordic Mathematical Contest)
Analysis: It is easy to see from observation that the left side of the equation can be factored.
Solution: \((x+y+z)\left[(x-y)^{2}+(y-z)^{2}+(z-x)^{2}\right]=4006\).
Since \(4006=2 \times 2003\) and \((x-y)^{2}+(y-z)^{2}+(z-x)^{2} \equiv 0(\bmod 2)\),
we have
\[
\left\{\begin{array}{l}
x+y+z=1 \\
(x-y)^{2}+(y-z)^{2}+(z-x)^{2}=4006
\end{array}\right.
\]

or
\[
\left\{\begin{array}{l}
x+y+z=2003 \\
(x-y)^{2}+(y-z)^{2}+(z-x)^{2}=2
\end{array}\right.
\]

For (1),
\[
(x-y)^{2}+(x+2 y-1)^{2}+(2 x+y-1)^{2}=4006.
\]

That is,
\[
6 x^{2}+6 y^{2}+6 x y-6 x-6 y+2=4006.
\]

But \(4006 \equiv 4(\bmod 6)\) leads to a contradiction.
For (2), since \(|x-y|,|y-z|,|z-x|\) are two 1s and one 0, without loss of generality, assume \(x \geqslant y \geqslant z\).
When \(x-1=y=z\), \(3 y+1=2003\) has no solution;
When \(x=y=z+1\), \(3 x-1=2003 \Rightarrow x=668\).
Therefore, the integer solutions satisfying the conditions are \((668,668,667),(668,667,668),(667,668,668)\).
/Comment: Factoring is one of the main methods for solving indeterminate equations.