Suppose $\alpha\in \mathbb{N}$, and
$$A=\{(x,y)\in \mathbb{N}^2 \mid x\ne y,\quad \alpha xy - 1|(\alpha x^2-1)^2 \}$$ is the relation $A$ symmetic?
When is it empty?
For more information please see my last post in Are $4ab\pm 1 $ and $(4a^2\pm 1)^2$ coprime?
This is a standard "vieta jumping" problem.
Manipulations yield $( \alpha ab-1 )|(\alpha a^{2}-1)^{2}$ implies $(\alpha ab - 1)|(\alpha a^2b - b)^2$ and thus $(\alpha ab - 1)|(a-b)^2$.
Let's assume now that there exist some postive integers $ a\neq b$ so that $ (\alpha ab-1)|(a-b)^{2}$. Let $\displaystyle k = \frac{(a-b)^{2}}{\alpha ab-1} > 0$.
Let $ S=\{ (a,b): a,b\in \mathbb{Z}^{+}, (a-b)^{2}/(\alpha ab-1)=k\}$. Since $ |S| \geq 1$, there is a pair $ (a,b)\in S$ which minimizes $ a+b$ over all $ (a,b)\in S$. WLOG assume that $a > b$. Consider now the quadratic equation $$ \frac{(x-b)^{2}}{\alpha xb-1}=k \Leftrightarrow x^{2}-x(2b+bk \alpha)+b^{2}+k=0 $$
which has roots $ x_{1}=a$ and $\displaystyle x_{2}=2b+\alpha kb-a= \frac{b^{2}+k}{a}$ (from Vieta). This implies that $ x_{2}\in \mathbb{Z}^{+}$, so $ (x_{2}, b)\in S$. By the minimality of $a+b$, we however have $ x_{2}\geq a$, ie $\displaystyle \frac{b^{2}+k}{a} \geq a$ and therefore $ k\geq a^{2}-b^{2}$.
Thus, $\frac{(a-b)^2}{\alpha ab - 1} =k \geq a^{2}-b^{2}$ and hence $ a-b\geq (a+b)(\alpha ab-1)$, clearly impossible for $a,b \in\mathbb{Z}^{+}$. Thus if the divisibility condition holds then $a=b$.
This relation is never empty. Indeed, it always contains any $(x,0)$ with $x \gt 0$ .
It is always symmetric. Indeed, if $(x,y)\in A$, then there is an integer $k$ such that $(ax^2-1)^2=k(axy-1)$.
Put
$$ q=(axy)^3+(axy)^2+(axy)-2(axy+1)(ay^2)+1 $$
Then ,we have the identity
$$ (ay^2)^2((ax^2-1))^2-(axy-1)q=((ay^2-1))^2 $$
and therefore $(ax^2-1)^2=k'(axy-1)$ with $k'=(ay^2)^2k-q$. So $(y,x)\in A$.
