I would like to solve if it is possible next diophantine equation for positive integers $n\geq 1$, $x\geq 1$ and $y\geq 1$
$$n x^2-\operatorname{rad}(n)=y^3,\tag{1}$$ where $\operatorname{rad}(n)$ denotes the product of distinc primes dividing our integer $n>1$ with the definition $\operatorname{rad}(1)=1$ (it is the arithmetic function Radical of an integer from Wikipedia).
Computational facts. Using a Pari/GP program the solutions, we write these as $(x,y,n)$, that I can to find in next segments $1\leq x\leq 500$, $1\leq y\leq 500$ and $1\leq n\leq 500$ are respectively $(x,y,n)=(3,2,1)$, $(17,12,6), (26,15,5),(99,70,35)$ and $(485,198,33)$. Thus I've no enough computational evidence to say if these are a complete set of solutions pf $(1)$.
Claim 0 (Preparation). For each fixed $n>1$ (see Claim 1) and being $\prod_{p\mid n}p^{e_p}$ the prime factorization of $n$ (fundamental theorem of arithmetic) we are searching/we need to characterize integers with the form $$\left(\prod_{p\mid n}p^{e_p-1}\right)\cdot x^2-1\tag{2}$$ and that are divisible by $p^2$ for each prime $p$ dividing $n$.
Question. I would like to know a full characterization of the solutions of our equation, or at least to know if there exist infinitely many or finitely many of such solutions. I wrote previous Claim 0 that I believe that is the key result that combined to Euler-Fermat theorem will provide us an inventory of solutions. I add also auxiliary claims that I think that will be useful to finish this task. Can you find the solutions of $$n x^2-\operatorname{rad}(n)=y^3,$$ over positive integers? Is there a finite number of such solutions $(x,y,n)$? Many thanks.
Claim 1. The case $n=1$ is Mihăilescu's theorem (see the for example the Wikipedia's article dedicated to Catalan's conjecture).
Claim 2. It's easy to prove that $n$ has no repeated prime factors.
Proof. By contradiction let $p\mid n$ such that $p^2\mid n$ then from $$n\cdot x^2-y^3=\operatorname{rad}(n)$$ we deduce the absurd that $p^2\mid \operatorname{rad}(n).$ We've used the obvious fact that for each prime $p$ dividing $n$ then $p^3\mid y^3$ holds$\square$.
Claim 3. One has that for each solution $(x,y,n)$ the condition $\gcd(x,n)=1$ is satisfied.
Proof. For the case $n=1$ we invoke again Mihăilescu's theorem. By contradiction, for $n>1$ we assume that there exists a prime $p\mid \gcd(x,n)$ and we get a contradiction using a similar argument than previous proof. $\square$
Partial result :
We want to show that the equation $$nx^2-n=y^3$$ has infinite many integer solutions with positive squarefree $n$ and positive integers $x,y$
The given equation is equivalent to $$x^2-\frac{y}{n}\cdot y^2=1$$
Assume $x,y$ is a solution of the pell-equation $$x^2-2y^2=1$$ and $\frac{y}{2}$ is a positive squarefree integer. Then with $n:=\frac{y}{2}$, we get the desired solution. Hence, $(1)$ has probably infinite many solutions.
The first solutions of this kind are :