Integer solutions of (x+1)(xy+1)=z^3

Question

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 $y$.

Answer 1

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 $x^2/2+O(x)$.

Answer 2

Here is another family of solutions.

For natural $q$:

$x=3,y=144 q^{3} + 360 q^{2} + 300 q + 83,z=12q+10$.

Yours is linear in $y$, so solutions are positive integers of the form $y=\frac{z^{3} - x - 1}{x^{2} + x}$.

Plugging small values of $x$, more solutions may come from integers:

x= 2   y= (1/6) * (z^3 - 3)
x= 3   y= (1/12) * (z^3 - 4)
x= 4   y= (1/20) * (z^3 - 5)
x= 5   y= (1/30) * (z^3 - 6)
x= 6   y= (1/42) * (z^3 - 7)
x= 7   y= (1/56) * (z - 2) * (z^2 + 2*z + 4)
x= 8   y= (1/72) * (z^3 - 9)
x= 9   y= (1/90) * (z^3 - 10)
x= 10   y= (1/110) * (z^3 - 11)
x= 11   y= (1/132) * (z^3 - 12)
x= 12   y= (1/156) * (z^3 - 13)
x= 13   y= (1/182) * (z^3 - 14)
x= 14   y= (1/210) * (z^3 - 15)