Example 7 Find all integer arrays $(a, b, c, x, y, z)$, such that
$$
\left\{\begin{array}{l}
a+b+c=x y z, \\
x+y+z=a b c .
\end{array}\right.
$$

Here $a \geqslant b \geqslant c \geqslant 1, x \geqslant y \geqslant z \geqslant 1$.
By symmetry, we only need to consider the case $x \geqslant a$. In this case, $x y z = a + b + c \leqslant 3 a \leqslant 3 x$, so $y z \leqslant 3$. Thus, $(y, z) = (1,1), (2,1), (3,1)$.

When $(y, z) = (1,1)$, $a + b + c = x$ and $x + 2 = a b c$, so $a b c = a + b + c + 2$. If $c \geqslant 2$, then $a + b + c + 2 \leqslant 3 a + 2 \leqslant 4 a \leqslant a b c$, with equality if and only if $a = b = c = 2$; if $c = 1$, then $a b = a + b + 3$, i.e., $(a-1)(b-1) = 4$, yielding $(a, b) = (5,2)$ or $(3,3)$.

When $(y, z) = (2,1)$, $2 a b c = 2 x + 6 = a + b + c + 6$. Similar discussion shows that $c = 1$, and thus $(2 a - 1)(2 b - 1) = 15$, yielding $(a, b) = (3,2)$.
When $(y, z) = (3,1)$, $3 a b c = 3 x + 12 = a + b + c + 12$, and it can be seen that there is no solution with $x \geqslant a$.
In summary, the solutions are $(a, b, c, x, y, z) = (2,2,2,6,1,1), (5,2,1,8,1,1), (3,3,1,7,1,1), (3,2,1,3,2,1), (6,1,1,2,2,2), (8,1,1,5,2,1)$, and $(7,1,1,3,3,1)$.