Cross-multiply. \[\frac{x+1}{x} \cdot \frac{y+1}{y}=\frac{z+1}{z}\implies(x+1)(y+1)z=xy(z+1)\implies xz+yz+z=xy\implies xy-xz-yz+z^2=z(z+1)\implies (x-z)(y-z)=z(z+1)\] For each integer $z$, this equation tells you to look at all factorizations of $z(z+1)$ as $a\cdot b$ and set $x=a+z$ and $y=b+z$. This gives all solutions.

It is equivalent to have $x+y+1 \vert x(x+1)$. So the solutions are $(x, d-x-1, x-\frac{x(x+1)}{d}$) and $(d-x-1,x,x-\frac{x(x+1)}{d})$ where $d>x+1$ is a divisor of $x(x+1)$.

$x+y\neq-1$

Rewrite to get $z(x+y+1)=xy$. Let $x=dx_1,y=dy_1$, and $z=dz_1$ where $d=gcd(x,y,z)$. We can substitute to get $z_1\left(dx_1+dy_1+1\right)=dx_1y_1$. Thus $d \mid z_1$. Let $z_1=dz_2$ to get $$z_2\left(dx_1+dy_1+1\right)=x_1y_1.$$Let $gcd\left(x_1,z_2\right)=a$ and $gcd\left(y_1,z_2\right)=b$. And as $gcd\left(x_1,y_1,z_2\right)=gcd\left(x_1,y_1,z_1\right)$ we have $gcd(a,b)=1$ thus $ab \mid z_2$. Now we can replace $x_1,y_1,z_2$ with $ax_2,by_2$, and $abz_3$ respectively. Thus substituting we have $z_3\left(dax_2+dby_2+1\right)=x_2y_2$ but this means $z_3 \mid x_2y_2$ however $gcd\left(z_3,x_2\right)=gcd\left(z_3,y_2\right)=1$ we get that $z_3=1$. Now we only need to solve $dax_2+dby_2+1=x_2y_2$ or $$\left(x_2-db\right)\left(y_2-da\right)=d^{2}ab+1.$$Set $y_2-da=t$, which gives $x_2=\frac{d^{2}ab+dbt+1}{t}$. Then the solutions are $$(x,y,z)=\left(da\cdot\frac{d^{2}ab+dbt+1}{t},db(da+t),d^{2}ab\right).$$