This post got me curious. The Diophantine equation with $x\neq1$,
$$x^n+y^3 = z^2\tag1$$
has infinitely many coprime solutions when $n\leq5$. Are there at least a few co-prime ones when $n>5$?
$n=6:$ $\color{red}{??}$
No solutions with $x<100$ and $0<y<100000$. (I checked.)
$n=7:$
$$2^7+17^3=71^2$$ $$17^7+76271^3=21063928^2$$
$n=8:$
$$43^8+96222^3 = 30042907^2$$
Q: Does $n=6$ in fact have a coprime solution? (It seems strange that $n=7,8$ has but $n=6$ doesn't. A larger search radius might yield a result.)
P.S.: From work by Darmon and Granville, if $1/p+1/q+1/r<1$, then the equation $ax^p+by^q+cz^r=0$ has only finitely many coprime solutions. So $(1)$ has only finitely many for $n\geq7$.
Proposition 14.6.1. In the parabolic case $1/p + 1/q + 1/r = 1$, the equation $x^p + y^q = z^r$ has no solutions in nonzero coprime integers, except that the equation $x^3 + y^6 = z^2$ has the solutions $(x, y, z) = (2, ±1, ±3)$, and the equation $x^3 + y^2 = z^6$ has the solutions $(x, y, z) = (−2, ±3, ±1)$.
From Henri Cohen Number Theory Volume II: Analytic and Modern Tools