Find all $x,y,z\in\mathbb N$, $x,y,z>1$ such that satisfy $x\mid yz+1$, $y\mid xz+1$, and $z\mid xy+1$

Find all $x,y,z\in\mathbb N$, $x,y,z>1$ such that satisfy $$\begin{cases}x\mid yz+1\\y\mid xz+1\\z\mid xy+1\end{cases}$$

I've found out easily that $$\begin{cases}x\nmid yz\\y\nmid xz\\z\nmid xy\end{cases}\Rightarrow \begin{cases}x\nmid y\\x\nmid z\\y\nmid x\\y\nmid z\\z\nmid x\\z\nmid y\end{cases}$$ I can't find a way to continue from here, so I'd like to get some ideas. Thanks.

The only solutions are permutations of $(2,3,7)$:

Assume $x \mid yz+1$, $y \mid xz+1$, $z \mid xy+1$. We have $(x,y)= (y,z) = (z,x) = 1$. Let $x< y< z$. Then as $x$,$y$ and $z$ divide $xy+xz+yz+1$ we have $$xyz \mid xy+xz+yz+1 < 3yz.$$ By $1<x$ we conclude $x= 2$. But then $y \mid (2z+1)$ and $z \mid (2y+1)$ leading to $$yz \mid 2(y+z)+1 <4z$$ from which we conclude $y=3$ using $x<y$. Finally $z \mid xy+1 = 7$ and indeed $(2,3,7)$ is a solution.