213. Solve the equation $x y z = x + y$ in natural numbers $x, y$ and $z$.

Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly.
$\triangle$ Suppose that $x \leq y$. Such a restriction can be imposed mainly in cases where the equation is symmetric with respect to $x$ and $y$. It does not spoil the problem, as it can be removed during further solving.

We get:

$$
x y z = x + y \leq y + y = 2 y, \quad x y z \leq 2 y
$$

from which

$$
x z \leq 2 \text{. }
$$

Let's consider all cases.

1) Let $x=1, z=1$. Then

$$
y = y + 1 \text{, }
$$

which is impossible.

2) Let $x=2, z=1$. We have:

$$
2 y = 2 + y, \quad y = 2 \text{. }
$$

We found a solution $(2 ; 2 ; 1)$.

![](https://cdn.mathpix.com/cropped/2024_05_21_0aa0e6ac168e80891953g-074.jpg?height=602&width=1330&top_left_y=256&top_left_x=383)

3) Let $x=1, z=2$. Then

$$
2 y = 1 + y, \quad y = 1
$$

We found another solution $(1 ; 1 ; 2)$.

Answer: $(2 ; 2 ; 1), (1 ; 1 ; 2)$.