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. 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 $t&lt;4$ 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 $z&gt;21,$ 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:

$(x,y,z,t)=(2,4, (-1),(-2))$
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:

$(x,y,z,t)=(2,4, (-1),(-2))$