Can that finiteness be proved for polynomials $P^n\neq\pm Q^m,\quad n,m>0\;$ by known methods?
For univariate polynomials $A,B,X,Y$ $$AX+BY\equiv0$$ iff $$A\equiv\frac{HY}{(X,Y)},B\equiv-\frac{HX}{(X,Y)}$$ Consider $\deg A<m=\deg Y,\deg B<n=\deg X$. Since $$AX+BY\equiv0 \\ \leftrightarrow \\ (AX+BY)(\alpha_i)=0,\quad i=1,\dots,n+m\ge\deg(AX+BY)+1$$ then representing $A$ in the Lagrange form on $$y_1=\alpha_1,\dots,y_m=\alpha_m$$ and $B$ in the Lagrange form on $$x_1=\alpha_{m+1},\dots,x_n=\alpha_{m+n}$$ one obtains homogeneous linear system with matrix of coefficients $$\begin{pmatrix} Y(x_1)\prod_{i=2}^n(x_1-x_i) & 0 & \dots & X(x_1)\prod_{i=2}^m(x_1-y_i) & X(x_1)(x_1-y_1)\prod_{i=3}^m(x_1-y_i) & \dots \\ \vdots \\ Y(y_1)\prod_{i=1}^n(y_1-x_i) & Y(y_1)(y_1-x_1)\prod_{i=1}^m(y_1-x_i) & \dots & X(y_1)\prod_{i=2}^m(y_1-y_i) & 0 & \dots \\ \end{pmatrix}$$ Determinant of this matrix equals to zero iff $(X,Y)=1$.
Finiteness of solutions of initial system is equivalent to $$(PP(y)-1,QQ(y)-1)=1$$ Let $$P(x)=p(x-p_1)\dots(x-p_\nu)(x-r_1)\dots(x-r_\lambda) \\ Q(x)=q(x-q_1)\dots(x-q_\mu)(x-r_1)\dots(x-r_\lambda)\neq const,$$ where $$p_i\neq q_j,p_i\neq r_j,q_i\neq r_j, \\ p_1\neq p_k,q_i\neq q_k,r_i\neq r_k$$ for $$i\neq k$$ and $$\nu>\mu$$ Substitution $$X=PP(y)-1,Y=QQ(y)-1 \\ x_1=p_1,\dots,x_\nu=p_\nu,x_{\nu+1}=r_1,\dots \\ y_1=q_1,\dots,y_\mu=q_\mu,y_{\mu+1}=a_1,\dots$$ ($a_i\neq p_j,a_i\neq q_j,a_i\neq r_j,a_i\neq a_k$ for $i\neq k$) yelds polinomial expression of determinant on y that have $$\deg=\nu(\mu+\lambda)+(\mu+\lambda)(\nu+\lambda)=(n-\lambda)m+mn=2mn-m\lambda$$
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Thus initial system has finite number of solutions for inconstant polinomials with different degrees and without multiple roots.