Example 7 Let $x, y, z$ be integers greater than 2, and
$$
\begin{array}{l}
x y \equiv 2(\bmod z), y z \equiv 2(\bmod x), \\
z x \equiv 2(\bmod y) .
\end{array}
$$

Find the value of $x+y+z$.
Three congruences are equivalent to
$$
x|(y z-2), y|(z x-2), z \mid(x y-2) \text {. }
$$

Therefore, $x y z \mid[(y z-2)(z x-2)(x y-2)]$.
Notice that
$$
\begin{array}{l}
(y z-2)(z x-2)(x y-2) \\
=A x y z+4(x y+y z+z x)-8 .
\end{array}
$$

Thus, $4(x y+y z+z x)-8=k x y z$, which means
$$
4\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)-\frac{8}{x y z}=k \text {. }
$$

Using the integers $x, y, z>2$, we can estimate the range of $k$,
$$
k1 \text {, }
$$

the equation cannot hold;
If $x \geqslant 12$, then
$$
k<4\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right) \leqslant 4\left(\frac{1}{12}+\frac{1}{13}+\frac{1}{14}\right)<1 \text {, }
$$

which is a contradiction. Therefore, $x=5,6, \cdots, 11$.
Similarly, by enumerating and discussing seven cases, we can obtain two sets of solutions that satisfy the system of congruences:
$$
(x, y, z)=(6,14,82),(6,22,26) .
$$

For each of the above solutions, calculate the value of $x+y+z$.