Divisibility of a general polynomial of a rational expression [closed]

Let $P_0(y),P_1(y),...,P_n(y)\in\mathbb{C}[y]$ where $P_n(y)\neq 0$, and let $p(x),q(x)\in\mathbb{C}[x]$ coprime, both nonzero and $q(x)$ nonconstant. Consider the multivariate polynomial $$P(x,y)=P_0(y)p(x)^0q(x)^n+P_1(y)p(x)^1q(x)^{n-1}+...+P_n(y)p(x)^nq(x)^0.$$ Can $q(x)$be a divisor of $P(x,y)$?

I already found that $q(x)$ cannot be a divisor of $p(x)$, $q(x)^0$ and $P_0(y),...,P_n(y)$.

The question comes from deciding under which conditions an equation $$\sum_{i=0}^nP_i(y)\left(\frac{p(x)}{q(x)}\right)^i=0$$ is irreducible if the polynomial $\sum_{i=0}^nP_i(y)x^i$ is irreducible. I want to be sure and need confirmation therefore.

Clearly $q(x)$ divides $P_k(y)p(x)^kq(x)^{n-k}$ for all $k<n$. So the question is whether $q(x)$ can divide $P_n(y)p(x)^n$. Given that $\gcd(p(x),q(x))=1$, the question is whether $q(x)$ can divide $P_n(y)$. Given that $P_n(y)\neq0$ and $q(x)$ is not constant, the answer is no.