<s>[INST] Integer solutions of (x+1)(xy+1)=z^3 Consider the equation
$$(x+1)(xy+1)=z^3,$$
where $x,y$ and $z$ are positive integers with $x$ and $y$ both at least $2$ (and so $z$ is necessarily at least $3$). For every $z\geq 3$, there exists the solution
$$x=z-1 \quad \text{and} \quad y=z+1.$$
My question is, if one imposes the constraint that
$$x^{1/2} \leq y \leq x^2 \leq y^4,$$
can there be any other integer solutions (with $x,y \geq 2$)? 
Moreover, $z$ may be assumed for my purposes to be even with at least three prime divisors.
Thank you in advance for any advice.
Edit: there was a typo in my bounds relating $x$ and $y$.
 [/INST] There are other solutions. Try $x+1=a^3$, $xy+1=b^3$, where $b$ is chosen so that $b^2+b+1$ is divisible by $a^3-1$. Such $b$ is always possible to choose provided that $a-1$ is not divisible by 3 and by primes of the form $3k-1$. Moreover, $b$ may be replaced to its remainder modulo $a^3-1=x$, in this case $xy<xy+1=b^3<x^3$, thus $y<x^2$. If $b<a^3/2-1$, replace $b$ to $a^3-b-2$, it makes $y$ of order at least $x^2/2+O(x)$.
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