Just multiplieng the conditions

$xyz\mid (x+1)(y+1)(z+1)=xyz+xy+yz+zx+x+y+z+1\rightarrow xyz\mid xy+yz+zx+x+y+z+1\rightarrow xyz\leq xy+yz+zx+x+y+z+1$...

First , if $x=1$ or $y=1$ or $z=1$. WLOG; $x=1$ $z\mid 2$ $z=1$ or $2$. $z=1$ then $y\mid 2$ $y=1$ or $2$ $z=2$ then $y\mid 3$ $y=1$ or $3$ Therefore, $(x,y,z)=(1,1,1),(1,2,1),(1,3,1)$ and their permutation. Now, if $x,y,z\geq 2$ Mahdi.sh wrote: $xyz\mid (x+1)(y+1)(z+1)=xyz+xy+yz+zx+x+y+z+1\rightarrow xyz\mid xy+yz+zx+x+y+z+1\rightarrow xyz\leq xy+yz+zx+x+y+z+1$... From Last Inequality that @above show. $yz+y+z+1\geq x(yz-y-z-1) \geq 2yz-2y-2z-2$ $yz-3y-3z-3\leq 0$ $(y-3)(z-3)\leq 12$ and then case work

Mathmick51 wrote: First , if $x=1$ or $y=1$ or $z=1$. WLOG; $x=1$ $z\mid 2$ $z=1$ or $2$. $z=1$ then $y\mid 2$ $y=1$ or $2$ $z=2$ then $y\mid 3$ $y=1$ or $3$ Therefore, $(x,y,z)=(1,1,1),(1,2,1),(1,3,1)$ and their permutation. Now, if $x,y,z\geq 2$ Mahdi.sh wrote: $xyz\mid (x+1)(y+1)(z+1)=xyz+xy+yz+zx+x+y+z+1\rightarrow xyz\mid xy+yz+zx+x+y+z+1\rightarrow xyz\leq xy+yz+zx+x+y+z+1$... From Last Inequality that @above show. $yz+y+z+1\geq x(yz-y-z-1) \geq 2yz-2y-2z-2$ $yz-3y-3z-3\leq 0$ $(y-3)(z-3)\leq 12$ and then case work Are you sure that $(1,3,1)$ is a solution?

After realizing my mistake, let me present my humble and lengthy solution. $\left\{\begin{array}{cc}x|y+1\\y|z+1\\z|x+1\end{array}\right.$ $\implies$ $xyz|(x+1)(y+1)(z+1)$ $\implies$ $xyz|xyz+xy+yz+zx+x+y+z+1$ $\implies$ $xyz|xy+yz+zx+x+y+z+1$. So, we are looking for all triples of positive integers such that $\frac{xy+yz+zx+x+y+z+1}{xyz}\in\mathbb{N}$ or $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}+\frac{1}{xyz}\in\mathbb{N}$. Let define a function on the set of natural numbers: $f(x,y,z):=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}+\frac{1}{xyz}$. I will not waste a time on the case when any two of the three are equal. So, assume that $x,y$ and $z$ are pairwisely different. Since $f(x,y,z)$ is a symmetric function (not the original conditions of the problem!), we can assume wlog $x<y<z$. $x\ge4.$ Then $y\ge5, z\ge6.$ $0<f(x,y,z)\le\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot4}+\frac{1}{4\cdot5\cdot6}=\frac{90}{120}<1$. So, in this case we don't get any solutions. $x=3.$ Then $y\ge4, z\ge5.$ $0<f(x,y,z)\le\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot3}+\frac{1}{3\cdot4\cdot5}=1$. So, we get only one solution: $(x,y,z)=(3,4,5)$ with its permutations. However, $(x,y,z)=(4,3,5)$ and its cyclic permutations satisfy the original problem. $x=2.$ Then $y\ge3, z\ge4.$ $0<f(2,y,z)\le\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot2}+\frac{1}{2\cdot3\cdot4}=\frac{36}{24}<2$. So, the only possibility is $f(2,y,z)=1$. $\implies$ $\frac{1}{2}+\frac{1}{y}+\frac{1}{z}+\frac{1}{2y}+\frac{1}{yz}+\frac{1}{z\cdot2}+\frac{1}{2yz}=1$ $\implies$ $(y-3)(z-3)=12=1\cdot12=2\cdot6=3\cdot4$ $\implies$ $(y,z)=(4,15),(5,9),(6,7)$. However, none of these satisfies the original problem. $x=1.$ Then $y\ge2, z\ge3.$ $1<f(1,y,z)\le\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot1}+\frac{1}{1\cdot2\cdot3}=\frac{18}{6}=3$. So, the only possibilities are $f(1,y,z)=2$ and $f(1,y,z)=3$. $f(1,y,z)=2$. $\implies$ $\frac{1}{1}+\frac{1}{y}+\frac{1}{z}+\frac{1}{1\cdot y}+\frac{1}{yz}+\frac{1}{z\cdot1}+\frac{1}{1\cdot yz}=2$ $\implies$ $(y-2)(z-2)=6=1\cdot6=2\cdot3$ $\implies$ $(y,z)=(3,8),(4,5)$. However, none of these satisfies the original problem. $f(1,y,z)=3$. $\implies$ $\frac{1}{1}+\frac{1}{y}+\frac{1}{z}+\frac{1}{1\cdot y}+\frac{1}{yz}+\frac{1}{z\cdot1}+\frac{1}{1\cdot yz}=3$ $\implies$ $(y-1)(z-1)=2=1\cdot2$ $\implies$ $(y,z)=(2,3)$. After checking, only $(3,2,1)$ and its cyclic permutations satisfy the original problem. So, all solutions are $(1,1,1), (1,2,1), (4,3,5), (3,2,1)$ and their cyclic permutations. $\blacksquare$

All solutions, with max $z$ are: $(x,y,z)=(1,1,1);(1,1,2);(2,1,3);(4,3,5)$

Arslan wrote: Now we can assume wlog $x<y<z$... This is not true since the condition is not symmetric in $x,y,z$, only cyclic. However, one can assume that either $x<y<z$ or $x>y>z$, if the three numbers are distinct. In the second case $x | y+1$ and $x \ge y+1 \Longrightarrow x=y+1$. Similarly, $y=z+1$. Then $z \mid x+1 = z+3$, hence $z \mid 3 \Longrightarrow z \in \{1, 3\}$. This gives the two additional solutions $(5,4,3)$, $(3,2,1)$ (and their cyclic permutations).