Determine all positive integer solutions for $\frac 1x + \frac 1y + \frac 1z + \frac 1{xy} + \frac1{yz} + \frac 1{xz} + \frac 1{xyz} = 1$ I need to determine all positive integer solutions for the equation: 
$$\frac 1x + \frac 1y + \frac 1z + \frac 1{xy} + \frac1{yz} + \frac 1{xz} + \frac 1{xyz} = 1.$$
This is how I have tried to do it: 
Mulitiplied both sides by $xyz$ to get
$$yz+xz+xy+z+x+y+1=xyz.$$
Factor it:
\begin{align}
x(y+z+1-yz)+yz+y+z &= -1   \\
x(y(1-z)+z+1)+y(1+z)+z &= -1
\end{align}
If $z=0$, we get
\begin{align}
x(y+1)+y=-1 &\iff xy+x+y=-1   \\
&\iff (x+1)(y+1)=0,
\end{align}
which gives us $x=y=-1$.
Is this all positive integer solutions? Or have I missed something?
EDIT: I am stupid. 
New attempt. 
If $z=1$, I get $2x+2y=-2 \iff x+y=-1$.
Then there is no solutions of positive integers for both $x$ and $y$ at the same time.  
If I try for $z=2$, I get
\begin{align}
x(y(1-2)+2+1)+y(1+2)+2=-1 &\iff x(3-y)+3y+2=-1   \\
&\iff 3x+3y+3-xy=0
\end{align}
and I won't get a solution where all the variables are positive integers. 

Rewrite the equation as
$$
z( - xy + x + y + 1) + (x+1)(y+1)=0.
$$
Suppose first that $xy=x+y+1$, i.e., $y(1-x)=x+1$. Then $x\neq 1$, so that $y=\frac{x+1}{1-x}$. Substituting this into the original equation gives
$$
x(x+1)=0,
$$
a contradiction. Now assume that $- xy + x + y + 1\neq 0$. Then
$$
z=\frac{(x+1)(y+1)}{xy-x-y-1}.
$$
When can this be a positive integer? For $x=1$ we have $z=-y-1<0$, a contradiction. For $x=2$ we have 
$$
z=\frac{3y+3}{y-3},
$$
which has solutions for $y=4,5, 6,7,9,15$. For $x=3$ we obtain
$$
z=\frac{2y+2}{y-2}.
$$
This is an integer for $y=3,4,5,8.$ For $x\ge 4$ it is easy to see that we have no solution in positive integers.
