Is $xz+1 $ a proper divisor of $a_3z^3+a_2z^2+a_1z+1$ finitely often? Given a polynomial $P=a_3z^3+a_2z^2+a_1z+1, z >0$ with non-negative integer coefficients $a_1, a_2, a_3\ne 0$, it appears if $P$ is not factorizable then there are finitely many positive integers $x, z$ such that $xz+1 \mid P(z)$, $xz+1<P(z)$. If $a_2=a_1=0$, the claim is true. The Diophantine equation $ (xz+1)(yz+1)=az^{3}+1$ has no solutions in positive integers with $z > a^2+2a$. However the proof for the general case doesn't follow directly from the proof for the case $P=a_3z^3+1$. Also for a particular triple $(a_1, a_2, a_3)$, what's the minimum value of $z$ such that $xz+1$ is not a proper divisor of  $P(z)$ for all $x>1$?
My thoughts:
If $P$ is factorizable then we can find integers $b_1, b_2, b_3$ such that $a_3z^3+a_2z^2+a_1z+1=(b_1z+1)(b_2z^2+b_3z+1)$. Expanding and comparing coefficients we get $b_1+b_3=a_1$, $b_1b_3+b_2=a_2$, and $b_1b_2=a_3$. Since $P$ is assumed non-factorizable, we will have to use this result somewhere in the proof.
 A: The conjecture is true. That is, if the integral cubic polynomial
$$P(Z)=a_3 Z^3+a_2 Z^2+a_1 Z+1$$
is irreducible in $\mathbb{Z}[Z]$ (hence also in $\mathbb{Q}[Z]$ by Gauss's lemma), then there are only finitely many positive integer solutions of the equation
$$(xz+1)(yz+1)=P(z).$$
1. First we consider the case when $x\mid a_3$ or $y\mid a_3$. By symmetry, it suffices to deal with the case $x\mid a_3$. We fix $x$ for this section. By long division, we get an integral quadratic polynomial $Q\in\mathbb{Z}[Z]$ and a nonzero integer $r\in\mathbb{Z}$ such that
$$a_3^2 P(Z)=(xZ+1)Q(Z)+r.$$
If $xz+1\mid P(z)$, then $xz+1\mid r$, hence there are finitely many possibilities for $z$ (and also for $y$).
2. Now we consider the case when $x\nmid a_3$ and $y\nmid a_3$. We rewrite the original equation as
$$tz=x+y-a_1\qquad\text{where}\qquad t:=a_3z+a_2-xy.$$
Here $t$ is an integer. If $t\leq 0$, then $x+y\leq a_1$, which leads to finitely many triples $(x,y,z)$. So let us focus on the case $t>0$. We use an identity inspired by the OP's earlier post:
\begin{align*}
(tx-a_3)(ty-a_3)&=t^2 xy-a_3 t(x+y)+a_3^2\\
&=t^2(a_3z+a_2-t)-a_3 t(tz+a_1)+a_3^2\\
&=-t^3+a_2 t^2-a_1 a_3 t+a_3^2.
\end{align*}
We conclude that $t\leq 3\max(|a_1|,|a_2|,|a_3|)$, for otherwise the LHS is positive, while the RHS is negative. Moreover, the factors on the LHS are nonzero integers by $x\nmid a_3$ and $y\nmid a_3$. So there are finitely many possibilities for the factors on the LHS (namely they are integral divisors of the finitely many possible values of the RHS), hence also for the triple $(x,y,z)$.
