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

$$x=z-1 \quad \text{and} \quad y=z+1.$$

I would like to know under what conditions there exist (or do not exist) other integer solutions with $x,y \geq 2$. More specifically, I am interested in the case where $x$ and $y$ are 'close' in the sense that $x^{1/2} \leq y \leq x^2 \leq y^4$. Moreover, $z$ may be assumed 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 $y$.

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 $y$.

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

$$x=z-1 \quad \text{and} \quad y=z+1.$$

I would like to know under what conditions there exist (or do not exist) other integer solutions with $x,y \geq 2$. More specifically, I am interested in the case where $x$ and $y$ are 'close' in the sense that $x^{1/2} \leq y \leq x \leq y^2$. Moreover, $z$ may be assumed to be even with at least three prime divisors.

Thank you in advance for any advice.

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

$$x=z-1 \quad \text{and} \quad y=z+1.$$

I would like to know under what conditions there exist (or do not exist) other integer solutions with $x,y \geq 2$. More specifically, I am interested in the case where $x$ and $y$ are 'close' in the sense that $x^{1/2} \leq y \leq x^2 \leq y^4$. Moreover, $z$ may be assumed 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 $y$.