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I believe there have been some changes in how exactly GIMPS operates in the past few years so the answer may be different from the one 6 years ago. I think rather than completely overwrite this question it would have been better to add a footnote in the question about the new data.
If $\omega$ is a primitive sixth root of unity, then for any $n=6k\pm 1$ you have $1^n+(\omega^2)^n=\omega^n$. This is believed to be essentially the only nontrivial persistent solution so you may wish to exclude it, but if not, that gives $FRM(6k\pm 1)=1$.
This is tangentially related at best but in case you are not familiar with it, you may be interested in anabelian geometry, which does manage to cover a lot of such higher dimensional "arithmetic" objects, although it is less concerned with representations of these groups but rather what information the groups themselves carry.
Of course you can scale the solutions by any integer. Faltings's theorem implies that this equation will have finitely many solutions up to scaling. That's probably about all you can say from just general principles.
Why this particular prime and this particular shape of a polynomial? Surely you didn't just randomly stumble upon them and randomly guessed that this holds. If there is anything in particular that lead you to these forms, it would be helpful to provide that context. The same applies to your other recent questions.
The left and right actions of negative matrices don't coincide, so you cannot just arbitrarily forget them on the right just because they act on the left too.
It should be the rectangular and the "isosceles tori", the latter being ones with a lattice basis of the form $\{1,\frac{1}{2}+ib\}$ for some $b\in\mathbb R$. However your homeomorphism should be correct, since the isomorphism with $\mathbb R^2\cong\mathbb C$ corresponds to taking $j$-invariant, and the invariant tori are the ones with real $j$-invariant.
You can probably realize $Y$ as an orbifold with $X$ its 2-sheeted covering in orbifold sense. But as you note, it will not be so as topological spaces. $Y$ will not be a manifold, it will have a "boundary" corresponding to tori isomorphic to their conjugates.
@FrancescoPolizzi It is actually very easy to see that any bijective map between (irreducible) curves is a homeomorphism, since any such curve is equipped with cofinite topology. So any example must be at least 2-dimensional.
