When is $(x-1)(y-1)(z-1)$ a factor of $xyz-1$? Let $x$, $y$, $z$ be three natural numbers such that $1< x< y< z$. For how many sets of values of $(x,y,z)$, is $(x-1)(y-1)(z-1)$ a factor of $xyz-1$?
I noticed that $(x-1)(y-1)(z-1)=(xyz-1)-z(x+y-1)-xy+x+y$.
But i don't know how to proceed from here. Any clues?

There are no other solutions besides $(2,4,8)$ and $(3,5,15)$, already
discovered in aRaRa's answer.
Let $f(x,y,z)=\frac{xyz-1}{(x-1)(y-1)(z-1)}$ for integers $z>y>x>1$. We want
to know when $k=f(x,y,z)$ is an integer. Note first that $k$ cannot equal $1$,
because this would imply $z=\frac{x+y-xy}{x+y-1}$ whence $xy\leq 1$ which is impossible.
Now
$$
f(x,y,y+1)-f(x,y,z)=\frac{(xy-1)(z-(y+1))}{y(x-1)(y-1)(z-1)}\geq 0 \tag{1}
$$
and
$$
f(x,x+1,x+2)-f(x,y,y+1)=\frac{t\bigg((2x^2+2x-1)t+(2x^3+4x^2-1)\bigg)}{(x-1)x(x+1)(y-1)y}\geq 0 \tag{2}
$$
where $t=y-(x+1)$, so that $f(x,y,z)\leq f(x,x+1,x+2)=\phi(x)=\frac{x^3 + 3*x^2 + 2*x - 1}{x^3 - x}$.
If $x\geq 4$, we have $f(x,x+1,x+2)=\frac{119}{60}-\frac{(x-4)(100+(x-1)(59x+15))}{60x(x^2-1)}$,
so $k\leq \frac{119}{60} <2$ and hence $k=1$ which is impossible as already shown above.
So $x$ can only be $2$ or $3$. As $\phi(2)=\frac{23}{6}<4$ and $\phi(3)=\frac{59}{24}<3$,
we have $k<4$ when $x=2$ and $k<3$ when $x=3$.
Note that $f(x,y,z)=k$ can be rewritten as
$$
z=\frac{k(x+y)-kxy-k+1}{k(x+y)-(k-1)xy-k}  \tag{3}
$$
When $x=2$ and $k=2$, (3) yields $z=\frac{3}{2}-y$ which is impossible.
When $x=2$ and $k=3$, (3) yields $z=\frac{3y-4}{y-3}$ which is possible iff $y=4$.
When $x=3$ and $k=2$, (3) yields $z=\frac{4y-5}{y-4}$ which is possible iff $y=5$.
This concludes the proof.
