I do not know whether there is any advantage to considering this problem in the ring of Eisenstein integers $\mathbb{Z}[\omega],$ where $\omega = e ^{\frac{2 \pi i}{3}},$ which is a PID. Then we have  to ask when we can have $(x - \omega y)(x +\omega y)(x - \omega^{2}y )(x+\omega^{2}y)$ dividing $(xy- \omega)(xy +\omega)(xy- \omega^{2})(xy +\omega^{2})$ in $\mathbb{Z}[\omega],$ where $x,y$ are rational integers. I have not been able to pursue this to provide further insight myself, but someone else might.

Later remark: It is easy to check that the power of $1-\omega$ dividing both expressions is the same: it is $0$ if $3$ divides $xy,$ and $2$ if $3$ does not divided $xy.$ Hence we can omit the prime  $1-\omega$ from our considerations, and we only need to worry about primes in $\mathbb{Z}[\omega]$ such that $N(\pi)$ is a rational prime congruent to $1$ (mod $3$). If $\pi$ is such a prime dividing the leftmost product, we note that $\pi$ divides exactly one of the terms in the rightmost product (and, in fact, $\pi$ also divides exactly one term in the leftmost product). This leads (if the required divisibility holds ) relatively easily to the observation (already made by the OP) that the leftmost expression divides $y^{12}-1$ (and/or $x^{12}-1,$ there is symmetry in $x$ and $y$), but it is unclear to me at present whether this viewpoint provides any more useful information.

Later edit: I noticed that Aaron Meyerowitz's observation (in a comment after Gerry Myerson's answer) that  if the required divisibility holds, then $x^{4} + x^{2}y^{2} +y^{4}$ divides $(x^{4}-1)(y^{4}-1)$ can be derived this way. That is not particularly surprising, and the direct derivation is easier. However, perhaps less obvious is that we also have that $x^{4} + x^{2}y^{2} +y^{4}$ divides $(y^{8}+y^{4}+1)(x^{8}+x^{4}+1)$. While 
$(x^{4} + x^{2}y^{2} +y^{4})^{2}$divides $(x^{12}-1)(y^{12}-1),$ it is not immediately obvious to me that this last claimed divisibility is a consequence of that- for example, there might a priori be a prime $\pi$ such that $x^{4}-1$ is divisible by some higher than expected power of $\pi$- so I outline a proof: 

Note that if $\pi$ is a prime in $\mathbb{Z}[\omega]$ with $N(\pi) \equiv 1$ (mod $3$), then if $\pi^{m}$ divides both $x^{4}-1$ and $x^{2}- \omega y^{2},$ we have $\pi^{m}$ divides $\omega^{2}y^{4}-1,$ so that $\pi^{m}$ divides $y^{4}-\omega^{4}.$ It follows that $N(\pi)^{m}$ divides $y^{8}+y^{4} + 1.$  Hence it follows that (in $\mathbb{Z}$), ${\rm gcd}(x^{4}-1,x^{4} + x^{2}y^{2} + y^{4})$ divides $y^{8}+y^{4}+1$ (as before, the power of $3$ is taken care of). Similarly ${\rm gcd}(y^{4}-1,x^{4} + x^{2}y^{2} + y^{4})$ divides $x^{8}+x^{4}+1$. Since $x^{4}+ x^{2}y^{2} + y^{4}$ divides $(x^{4}-1)(y^{4}-1),$ the claim is established.