Hyperelliptic curves imply FLT-like results Probably this is known, but doesn't show in searches.
If a certain hyperelliptic curve has only trivial rational points,
FLT-like curve also has only trivial rationals points for fixed $n$.
Working over the rationals for integer $a$ define:
$$ u^n-v^n=a \qquad (1)$$
From $(u^n-v^n)^2 + 4 (uv)^n=(u^n+v^n)^2$ we get
$$ x^n + a^2/4=y^2 \qquad (2)$$
From $(uv)^n=v^n(v^n+a)$ we get
$$ x^n=y(y+a) \qquad (3)$$
Non trivial rational point on (1) leads to non trivial $x \ne 0$ 
rational point on (2) and (3). The converse need not hold.
Only trivial points on the hyperelliptic curves means only
trivial on (1).
On MO answers find all rational points on a given hyperelliptic
with the help of computer.
Is finding the rational points on the hyperelliptic curves easier
than on the FLT-like curve?
Fix $a$. Is it feasible to find all rational points on the hyperelliptic curves
for $n$ up to say $50$?
Added 1 Is it known or open that for rational $y$, the solutions to $y(a+y)=x^k$ are finite for $k>4$ and nonzero $x$? Basically quadratic is perfect power at rationals
 A: Let $p$ be an odd prime, and assume $ax^p+by^p+cz^p=0$ with  $x\ne0$; then
$$Y^2=X^p+a^2(bc)^{p-1}/4$$
has a nontrivial point with
$$X=-bcyz/x^2\quad\,\quad Y=(-bc)^{(p-1)/2}(by^p-cz^p)/(2x^p)\;.$$
A: A reference for all quotients of the Fermat curve is Lang's book "Introduction to Algebraic and Abelian Functions". There, you'll find your maps (up to twist) and several others.
If you have a map $X \to Y$ and $X$ has a rational point, so does $Y$. At first sight, you'd think it would be easier to prove that $X$ has no points than $Y$ (assuming that's the case). But, in some situations, showing that $Y$ has no points can be easier. A famous instance is Mazur's work on modular curves (and torsion on elliptic curves), where he shows that certain quotients of modular curves have Mordell-Weil rank zero and thus a describable set of rational points. Maybe this is possible with the Fermat curves also, I don't know. Another possibility is to use Chabauty's method if the Mordell-Weil rank of $Y$ is less than its genus. These days, Chabauty's method is almost automated for hyperelliptic curves of low genus.
