For $x,y\ge 2$ does $x^4+x^2y^2+y^4$ ever divide $x^4y^4+x^2y^2+1$? For a problem in combinatorics, it comes down to knowing whether there exist integers $x,y\ge 2$ such that 
$$ 
x^4+x^2y^2+y^4\mid x^4y^4+x^2y^2+1.
$$
Note that $x^6(x^2-y^2)(x^4+x^2y^2+y^4)+(x^2y^2-1)(x^4y^4+x^2y^2+1)=x^{12}-1$ and so we can look at a perhaps simpler problem: Are the only solutions to 
$$ 
x^4+x^2y^2+y^4\mid x^{12}-1
$$
(where $x,y\ge 2$) given by $(x,y)=(5,6)$ and $(x,y)=(6,5)$?

It seems to me that you need both $$p=x^2+xy+y^2$$ and $$q=x^2-xy+y^2$$ to divide $$r=x^4y^4+x^2y^2+1$$ Considering all expressions as polynomials in $x$, the remainder when you divide $r$ by $p$ is $$(y^7-y^3)x-y^4+1$$ and the remainder when you divide $r$ by $q$ is  $$(y^3-y^7)x-y^4+1$$ if Maple and I are on the same page. These remainders are both zero if and only if $y=\pm1$. 
