Example 4 Let $x, y, z \in \mathbf{R}^{+}$, and $x y z(x+y+z)=1$, find the minimum value of $(x+y)(x+z)$.
From $x y z(x+y+z)=1$, we can get $x^{2}+x y+x z=\frac{1}{y z}$, thus
$$\begin{aligned}
(x+y)(x+z) & =x^{2}+x y+x z+y z=y z+\frac{1}{y z} \\
& =\left(\sqrt{y z}-\frac{1}{\sqrt{y z}}\right)^{2}+2,
\end{aligned}$$

It is easy to derive from equation (8) that the minimum value is 2.