Example 1 Find the integer solutions of the indeterminate equation $\frac{1}{2}(x+y)(y+z)(z+x)+(x+y+z)^{3}=1-$ $x y z$.

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Let's make the substitution, set $x+y=u, y+z=v, z+x=w$, then the original equation transforms to $4 u v w+(u+v+w)^{3}=8-(u+v-w)(u-v+w)(-u+v+w)$.
After expanding and combining like terms, we get
$$4\left(u^{2} v+v^{2} w+w^{2} u+u v^{2}+v w^{2}+w u^{2}\right)+8 u v w=8,$$

which simplifies to
$$u^{2} v+v^{2} w+w^{2} u+u v^{2}+v w^{2}+w u^{2}+2 u v w=2 \text {. }$$

Factoring the left side of the equation, we get
$$(u+v)(v+w)(w+u)=2,$$

Thus, $(u+v, v+w, w+u)=(1,1,2),(-1,-1,2),(-2,-1,1)$ and their symmetric cases. Solving each case, we obtain
$$(u, v, w)=(1,0,1),(1,-2,1),(-1,0,2)$$

Consequently, $(x, y, z)=(1,0,0),(2,-1,-1)$.
In summary, considering symmetry, the integer solutions to the original equation are
$$\begin{aligned}
(x, y, z)= & (1,0,0),(0,1,0),(0,0,1),(2,-1,-1), \\
& (-1,2,-1),(-1,-1,2)
\end{aligned}$$

There are 6 solutions in total.