Find all non-zero natural numbers $x, y$ and $z$ such that

$$
\left(1+\frac{1}{x}\right)\left(1+\frac{1}{y}\right)\left(1+\frac{1}{z}\right)=2
$$
Let $(x, y, z)$ be a potential solution triplet. Without loss of generality, we assume below that $x \leqslant y \leqslant z$. We first deduce that

$$
\left(1+\frac{1}{x}\right)^{3} \geqslant\left(1+\frac{1}{x}\right)\left(1+\frac{1}{y}\right)\left(1+\frac{1}{z}\right)=2=\frac{128}{64}>\frac{125}{64}=\left(1+\frac{1}{5}\right)^{3}
$$

so that $x \leqslant 4$. Moreover, we know that

$$
1+\frac{1}{1}=2>1+\frac{1}{x}
$$

so that $2 \leqslant x$. We therefore study the three plausible values of $x$.

- If $x=2$, we then verify that

$$
\left(1+\frac{1}{y}\right)^{2} \geqslant\left(1+\frac{1}{y}\right)\left(1+\frac{1}{z}\right)=\frac{4}{3}=\frac{196}{147}>\frac{192}{147}=\frac{64}{49}=\left(1+\frac{1}{7}\right)^{2}
$$

so that $y \leqslant 6$. Furthermore, the equality in the statement rewrites as

$$
z=\frac{3(y+1)}{y-3}
$$

so that $4 \leqslant y$. The solutions such that $x=2$ are therefore the triplets $(2,4,15),(2,5,9)$ and $(2,6,7)$.

- If $x=3$, we this time verify that

$$
\left(1+\frac{1}{y}\right)^{2} \geqslant\left(1+\frac{1}{y}\right)\left(1+\frac{1}{z}\right)=\frac{3}{2}=\frac{75}{50}>\frac{72}{50}=\frac{36}{25}=\left(1+\frac{1}{5}\right)^{2}
$$

so that $y \leqslant 4$. Furthermore, the equality in the statement rewrites as

$$
z=\frac{2(y+1)}{y-2}
$$

The solutions such that $x=3$ are therefore the triplets $(3,3,8)$ and $(3,4,5)$.

- If $x=4$, we finally verify that

$$
\left(1+\frac{1}{y}\right)^{2} \geqslant\left(1+\frac{1}{y}\right)\left(1+\frac{1}{z}\right)=\frac{8}{5}=\frac{40}{25}>\frac{36}{25}=\left(1+\frac{1}{5}\right)^{2}
$$

so that $y=4$. Furthermore, the equality in the statement rewrites as

$$
z=\frac{5(y+1)}{3 y-5}=\frac{25}{7}
$$

There is therefore no solution such that $x=4$.

In conclusion, the solutions, up to the order of the variables $x, y$ and $z$, are $(2,4,15),(2,5,9)$, $(2,6,7),(3,3,8)$ and $(3,4,5)$.