By symmetry, let's assume $x\ge y \ge z$. Then we have $$\Big(1+\frac{1}{z}\Big)^3 \ge \Big(1+\frac{1}{x}\Big)\Big(1+\frac{1}{y}\Big)\Big(1+\frac{1}{z}\Big)=2$$ This implies $z \le 3$. Do casework:

• $\underline{(1)\ z=1}$ This implies $\big(1+\frac{1}{x}\big)\big(1+\frac{1}{y}\big) =1$. This cannot hold because $\big(1+\frac{1}{x}\big) > 1$ and $\big(1+\frac{1}{y}\big)>1$ means $\big(1+\frac{1}{x}\big)\big(1+\frac{1}{y}\big) > 1$.
• $\underline{(2)\ z=2}$ This implies $\big(1+\frac{1}{x}\big)\big(1+\frac{1}{y}\big) =\frac{4}{3}$. Because $x\ge y$, therefore $$\Big(1+\frac{1}{y}\Big)^2\ge\Big(1+\frac{1}{x}\Big)\Big(1+\frac{1}{y}\Big)=\frac{4}{3}\implies 2\le y \le 6$$ Setting $y=2, 3, 4, 5$, and $6$, respectively, finds $(x, y, z) = (15, 4, 2), (9, 5, 2)$ are solutions.
• $\underline{(3)\ z=3}$ This implies $\big(1+\frac{1}{x}\big)\big(1+\frac{1}{y}\big) =\frac{3}{2}$. By a similar reasoning, we have $$\Big(1+\frac{1}{y}\Big)^2\ge\Big(1+\frac{1}{x}\Big)\Big(1+\frac{1}{y}\Big)=\frac{3}{2}\implies 3\le y \le 4$$ Setting $y=3$ and $4$ respectively finding $(x, y, z)= (8, 3,3)$ and $(5, 4, 3)$ are solutions.

Therefore, the solutions are all the permutations of $\boxed{(7, 6, 2), (9, 5, 2), (15, 4, 2), (8, 3, 3), (5, 4, 3)}$.