One possible fix is as follows: let us write $(\deg_X(E),\deg_Y(E))=(a,b)$ and $(\deg_X(P),\deg_Y(P))=(c,d)$. Let us assume $a<c$ and $b<d$ and $E$ coprime with $P$ in the factorial ring $k[X,Y]$. Suppose we have $x_1,\dots,x_n$ distinct points in $k$ such that $E(x_i,Y)\mid P(x_i,Y)$ in $k[Y]$.

The usual matrix $M(E,P)\in M_n(k[X])$ with coefficients in $k[X]$ whose determinant is the resultant $\mathrm{Res}_Y(E,P)\in k[X]$ is the matrix of the linear map $$ \phi = \phi_{E,P}: \left\{ \begin{array}{ccc} K_{<d}[Y]\times K_{<b}[Y] & \longrightarrow & K_{<b+d}[Y]\\ (U,V) & \longmapsto & UE+VP \end{array} \right. $$ where $K=k(X)$.

Suppose $\mathbf{E_b(x_i)\neq 0}$: then $E(x_i,Y)$ is a degree $b$ polynomial in $Y$ and the quotient $E(x_i,Y)Q_i=P(x_i,Y)$ has $$\begin{array}{rcl} \deg_Y(Q_i) & = & \deg(P(x_i,Y))-\deg(E(x_i,Y))\\ & = & \deg(P(x_i,Y))-\deg_Y(E(X,Y))\\ & \leq & \deg_Y(P)-\deg_Y(E) = d-b \end{array}$$ and thus we can write $$ \phi(Y^\ell Q_i, -Y^\ell) = E(x_i,Y)\cdot Y^\ell Q_i - Y^\ell P(x_i,Y) =0 $$ for $\ell=0,\dots,b-1$, we have $0\leq \deg(Y^\ell Q_i)<b+\deg(Q_i)\leq d$ and $\deg(-Y^\ell)<b$ so that these are (linearly independent vectors) of the kernel of $M(E,P)(x_i)$. It is easy to deduce from that (the fact that the kernel of $M(E,P)(x_i)$ has dimension (exactly) $b$) that $x_i$ is a root of multiplicity at least $b$ of the resultant $\mathrm{Res}_Y(E,P)$.

Also $\deg\big(\mathrm{Res}_Y(E,P)\big)\leq ad+bc$. Since $E_b$ has degree $\leq \deg_X(E)=a$, then if there are $m>a$ different $x_i$ satisfying the divisibility property, then $\mathrm{Res}_Y(E,P)$ has at least $m-a$ distinct roots of multiplicity at least $b$, so if $$ (m-a)b>ad+bc $$ then $\mathrm{Res}_Y(E,P)=0$ and we have reached a contradiction.

One possible fix is as follows: let us write $(\deg_X(E),\deg_Y(E))=(a,b)$ and $(\deg_X(P),\deg_Y(P))=(c,d)$. Let us assume $a<c$ and $b<d$ and $E$ coprime with $P$ in the factorial ring $k[X,Y]$. Suppose we have $x_1,\dots,x_n$ distinct points in $k$ such that $E(x_i,Y)\mid P(x_i,Y)$ in $k[Y]$.

The usual matrix $M(E,P)\in M_n(k[X])$ with coefficients in $k[X]$ whose determinant is the resultant $\mathrm{Res}_Y(E,P)\in k[X]$ is the matrix of the linear map $$ \phi = \phi_{E,P}: \left\{ \begin{array}{ccc} K_{<d}[Y]\times K_{<b}[Y] & \longrightarrow & K_{<b+d}[Y]\\ (U,V) & \longmapsto & UE+VP \end{array} \right. $$ where $K=k(X)$.

Suppose $\mathbf{E_b(x_i)\neq 0}$: then $E(x_i,Y)$ is a degree $b$ polynomial in $Y$ and the quotient $E(x_i,Y)Q_i=P(x_i,Y)$ has $$\begin{array}{rcl} \deg_Y(Q_i) & = & \deg(P(x_i,Y))-\deg(E(x_i,Y))\\ & = & \deg(P(x_i,Y))-\deg_Y(E(X,Y))\\ & \leq & \deg_Y(P)-\deg_Y(E) = d-b \end{array}$$ and thus we can write $$ \phi(Y^\ell Q_i, -Y^\ell)(x_i) = E(x_i,Y)\cdot Y^\ell Q_i - Y^\ell P(x_i,Y) =0 $$ for $\ell=0,\dots,b-1$, we have $0\leq \deg(Y^\ell Q_i)<b+\deg(Q_i)\leq d$ and $\deg(-Y^\ell)<b$ so that these are (linearly independent vectors) of the kernel of $M(E,P)(x_i)$. It is easy to deduce from that (the fact that the kernel of $M(E,P)(x_i)$ has dimension (exactly) $b$) that $x_i$ is a root of multiplicity at least $b$ of the resultant $\mathrm{Res}_Y(E,P)$.

Also $\deg\big(\mathrm{Res}_Y(E,P)\big)\leq ad+bc$. Since $E_b$ has degree $\leq \deg_X(E)=a$, then if there are $m>a$ different $x_i$ satisfying the divisibility property, then $\mathrm{Res}_Y(E,P)$ has at least $m-a$ distinct roots of multiplicity at least $b$, so if $$ (m-a)b>ad+bc $$ then $\mathrm{Res}_Y(E,P)=0$ and we have reached a contradiction.

One possible fix is as follows: let us write $(\deg_X(E),\deg_Y(E))=(a,b)$ and $(\deg_X(P),\deg_Y(P))=(c,d)$. Let us assume $a<c$ and $b<d$ and $E$ coprime with $P$ in the factorial ring $k[X,Y]$. Suppose we have $x_1,\dots,x_n$ distinct points in $k$ such that $E(x_i,Y)\mid P(x_i,Y)$ in $k[Y]$.

The usual matrix $M(E,P)\in M_n(k[X])$ with coefficients in $k[X]$ whose determinant is the resultant $\mathrm{Res}_Y(E,P)\in k[X]$ is the matrix of the linear map $$ \phi = \phi_{E,P}: \left\{ \begin{array}{ccc} K_{<d}[Y]\times K_{<b}[Y] & \longrightarrow & K_{<b+d}[Y]\\ (U,V) & \longmapsto & UE+VP \end{array} \right. $$ where $K=k(X)$.

Suppose $\mathbf{E_b(x_i)\neq 0}$: then $E(x_i,Y)$ is a degree $b$ polynomial in $Y$ and the quotient $E(x_i,Y)Q_i=P(x_i,Y)$ has $$\begin{array}{rcl} \deg_Y(Q_i) & = & \deg(P(x_i,Y))-\deg(E(x_i,Y))\\ & = & \deg(P(x_i,Y))-\deg_Y(E(X,Y))\\ & \leq & \deg_Y(P)-\deg_Y(E) = d-b \end{array}$$ and thus we can write $$ \phi(Y^\ell Q_i, -Y^\ell) = E(x_i,Y)\cdot Y^\ell Q_i - Y^\ell P(x_i,Y) =0 $$ for $\ell=0,\dots,b-1$, and these are independent vectors of the kernel. It is easy to deduce from that (the fact that the kernel of $M(E,P)(x_i)$ has dimension (exactly) $b$) that $x_i$ is a root of multiplicity at least $b$ of the resultant $\mathrm{Res}_Y(E,P)$.

Also $\deg\big(\mathrm{Res}_Y(E,P)\big)\leq ad+bc$. Since $E_b$ has degree $\leq \deg_X(E)=a$, then if there are $m>a$ different $x_i$ satisfying the divisibility property, then $\mathrm{Res}_Y(E,P)$ has at least $m-a$ distinct roots of multiplicity at least $b$, so if $$ (m-a)b>ad+bc $$ then $\mathrm{Res}_Y(E,P)=0$ and we have reached a contradiction.

One possible fix is as follows: let us write $(\deg_X(E),\deg_Y(E))=(a,b)$ and $(\deg_X(P),\deg_Y(P))=(c,d)$. Let us assume $a<c$ and $b<d$ and $E$ coprime with $P$ in the factorial ring $k[X,Y]$. Suppose we have $x_1,\dots,x_n$ distinct points in $k$ such that $E(x_i,Y)\mid P(x_i,Y)$ in $k[Y]$.

The usual matrix $M(E,P)\in M_n(k[X])$ with coefficients in $k[X]$ whose determinant is the resultant $\mathrm{Res}_Y(E,P)\in k[X]$ is the matrix of the linear map $$ \phi = \phi_{E,P}: \left\{ \begin{array}{ccc} K_{<d}[Y]\times K_{<b}[Y] & \longrightarrow & K_{<b+d}[Y]\\ (U,V) & \longmapsto & UE+VP \end{array} \right. $$ where $K=k(X)$.

Suppose $\mathbf{E_b(x_i)\neq 0}$: then $E(x_i,Y)$ is a degree $b$ polynomial in $Y$ and the quotient $E(x_i,Y)Q_i=P(x_i,Y)$ has $$\begin{array}{rcl} \deg_Y(Q_i) & = & \deg(P(x_i,Y))-\deg(E(x_i,Y))\\ & = & \deg(P(x_i,Y))-\deg_Y(E(X,Y))\\ & \leq & \deg_Y(P)-\deg_Y(E) = d-b \end{array}$$ and thus we can write $$ \phi(Y^\ell Q_i, -Y^\ell) = E(x_i,Y)\cdot Y^\ell Q_i - Y^\ell P(x_i,Y) =0 $$ for $\ell=0,\dots,b-1$, we have $0\leq \deg(Y^\ell Q_i)<b+\deg(Q_i)\leq d$ and $\deg(-Y^\ell)<b$ so that these are (linearly independent vectors) of the kernel of $M(E,P)(x_i)$. It is easy to deduce from that (the fact that the kernel of $M(E,P)(x_i)$ has dimension (exactly) $b$) that $x_i$ is a root of multiplicity at least $b$ of the resultant $\mathrm{Res}_Y(E,P)$.

Also $\deg\big(\mathrm{Res}_Y(E,P)\big)\leq ad+bc$. Since $E_b$ has degree $\leq \deg_X(E)=a$, then if there are $m>a$ different $x_i$ satisfying the divisibility property, then $\mathrm{Res}_Y(E,P)$ has at least $m-a$ distinct roots of multiplicity at least $b$, so if $$ (m-a)b>ad+bc $$ then $\mathrm{Res}_Y(E,P)=0$ and we have reached a contradiction.