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