Show that the equation $x^2+y^2+z^2=x^2y^2$ has no integer solution,except $x=y=z=0$ Show that the equation $x^2+y^2+z^2=x^2y^2$ has no integer solution,except $x=y=z=0.$

Let one of the $x,y,z$ be even number.Let $x=2p$
$x^2+y^2+z^2=x^2y^2$
This gives $y^2+z^2$ is also even,which means either $y,z$ are both even or $y,z$ are both odd.
So either $x=2p,y=2q,z=2r$ is the solution or $x=2p,y=2q+1,z=2r+1$ is the solution.I put $x=2p,y=2q+1,z=2r+1$ in the equation $x^2+y^2+z^2=x^2y^2$ and see that it does not satisfy both sides,so it is not the solution.
When I put $x=2p,y=2q,z=2r$,i get $p^2+q^2+r^2=4p^2q^2$
I am confused here how to prove that $p^2+q^2+r^2=4p^2q^2$ does not hold true.
Also i cannot solve when $x$ is odd which is the second case.
Is my method correct?Please help me complete the solution.

This is a Pythagorean quadruple problem.there is a good way to prove that the only solution is (0,0,0)
assuming the equation is $x^2+y^2+z^2=t^2$ ,∀  t=xy
all integer positive solution given by 
$x=(l^2+m^2-n^2)/n$ , $y=2l$, $z=2m$ ,and $t=(l^2+m^2+n^2)/n$
Then $2l(l^2+m^2-n^2)=l^2+m^2+n^2$
one can find that $(l^2+m^2-n^2)(2l-1)=0$
suppose $2l-1=0$  refused
then $l^2+m^2-n^2=0 ⇒x=0$ ,and $xy=2n$
but $xy=0 ⇒ n=0 ⇒ l^2+m^2=0 ⇒ y=0 ,z=0$
therefore the only solution is (0,0,0)
