Prove that the equation $$xyzt=2\, xy +2xz +2xt +2yz +2yt +2zt +4\,x+4\,y+4\,z+4\,t+4$$ has finite numbers of integer solutions.
My attempt. Suppose that $x \leq y \leq z \leq t$ and $t=\max(t,4)$ ( the case $t<4$ is trivial). Then $$ xyzt\leq 2\, xt +2xt +2xt +2yt +2yt +2zt +4\,t+4\,t+4\,t+4\,t+t=6xt+4yt+2 zt+21t, $$ and we obtain the inequality $xyz\leq 6x+4y+2z+21$.
Then by the same idea, assuming $z>21,$ I get that $$xyz\leq 33z \implies xy\leq 33,$$ so, there are only finite integer solutions for $x,y.$ Now, for any fixed $x,y$ there are only finite numbers integer solutions for $z$ of the inequality $xyz\leq 6x+4y+2z+21$. Then the same for $t.$
Is it ok? Is there any more elegant solution?
Equation above shown below:
$xyzt=2\, xy +2xz +2xt +2yz +2yt +2zt +4\,x+4\,y+4\,z+4\,t+4$
Above has numerical solution:
$(x,y,z,t)=(2,4, (-1),(-2))$