This is hard, so I am looking for partial results and how hard it is.


Let $n>4$. Is it true that the hyperelliptic curve $x^n=y(y+1)$
doesn't have rational point with $x \ne 0$?


If necessarily assume $n$ is prime.


Integral points on the curve are heavily studied.


It is one of the simplest exponential diophantine equations over the
rationals.


What tools are used for solving it?


FLT implies there are no solutions with $y$ being $n$-th power.


**Added** No solution for given $n$ imply Fermat Last Theorem for $n$ as discussed [here](https://mathoverflow.net/questions/309091/hyperelliptic-curves-imply-flt-like-results) see (3) with a=1


Assume $u^n-v^n=1$ is counterexample to FLT. Then $(uv)^n=v^n(v^n+1)$ and $(uv,v^n)$ is non trivial rational point on the curve.



Rational perfect power values of $y(y+1)$
There are no such solutions. Let $x=a/b$ and $y=c/d$ be reduced fractions. Then 
$a^n/b^n=(c(c+d))/d^2$ and since both sides are reduced fractions we get that
$a^n=c(c+d)$ and $b^n=d^2$. From the first equation we deduce that $c=e^n$ and $c+d=f^n$ since $c$ and $c+d$ are co-prime and their product is an $n$-th power. 
Then $d=f^n-e^n$ so $(f^n-e^n)^2=b^n$. 
If $n$ is even write $n=2k$ and take square root to get $f^{2k}-e^{2k}=\pm b^k$. 
If the sign is $+$ then we get $f^{2k}=e^{2k}+b^k$ and since $k>2$ this implies by FLT that one of $e,b,f$ is 0. If $f=0$ then $y=-1$ so $x=0$. $b=0$ is certainly impossible. Finally, $e=0$ means $y=0$. So this option is excluded.
If the sign is $-$ we get $e^{2k}=f^{2k}+b^k$ and the same reasoning works.


If $n$ is odd, from $(f^n-e^n)^2=b^n$ we deduce that $b$ is a square so $b=g^2$ and then we can extract the root to get $f^n-e^n= \pm g^n$ and proceed in a way similar to the case where $n$ is even. 


