Forbidden polynomial identities by the abc conjecture

Question

The Mason–Stothers theorem states

Let $a(t), b(t)$, and $c(t)$ be relatively prime polynomials such that $a + b = c$, with coefficients that are either real numbers or complex numbers. Then $\max\{\deg(a),\deg(b),\deg(c)\} \le \deg(\operatorname{rad}(abc))-1$

It is generalized to multivariate polynomials. The bound is tight for 3 polynomials in the sense that there are examples $\deg(c)=\deg(\operatorname{rad}(abc)) - 1$.

The abc conjecture for integers appears to forbid some polynomial identities for multivariate polynomials that are seemingly allowed by the polynomial abc theorems.

Let $a,b,c$ be relatively prime polynomials with integer coefficients in $x,y$ and $a+b=c$.

Suppose $\deg(c) = \deg(\operatorname{rad}(abc)) - 1 $ and $x y$ divides $\operatorname{rad}(abc)$. Setting $x,y$ to large coprime integer powers in practice will make $x y$ vanish in the radical, $c = O(x^{(\operatorname{rad}(a(x,y)b(x,y)c(x,y)) - 1)})$ while $\operatorname{rad}(abc)=O(x^{(\operatorname{rad}(a(x,y)b(x,y)c(x,y)) - 2)})$ in a sense decreasing the "degree" of the radical by $2$ and giving an infinite family of counterexamples to abc.

Another forbidden identity appears to occur if $\deg(c) = \deg(\operatorname{rad}(abc)) - 2 $ and $ x y (x^2-y^2-1) $ divides $\operatorname{rad}(abc)$ and $ (x^2-y^2-1) $ doesn't divide $c$. In $ x y (x^2-y^2-1) $ setting $x,y$ to consecutive powerful numbers $s+1,s$ gives an upper bound for the radical $\operatorname{rad}(2 s (s+1)) \sim s = y$ which reduces the "degree" of the radical by $3$, again contradicting abc.

Another one appears to occur if $\deg(c) = \deg(\operatorname{rad}(abc)) - 1 $ and $x^4 + k y^4$ divides $abc$ and the elliptic curve $u^4 + k = v^2$ is of positive rank.

I suppose the upper bound for the radical is OK in both cases, but not sure if the estimate for $c$ is correct -- some cancellation might make $c$ smaller, that is why the restriction on divisibility of $c$.

(A potential problem might be $a,b$ not being coprime in integers, but clearing the gcd might solve that, not entirely sure about this).

Question:

Does the abc conjecture really imply the above identities don't exist? (My reasoning might well be wrong).

Subquestion in case the main question is true:

Is there an unconditional proof that these polynomial identities don't exist?

EDIT: In comments Peter Mueller claimed that the slight modification of the second identity $x y (x^2-y^2) $ divides $ a b $, $\deg(c)=\deg(\operatorname{rad}(abc)) - 2$ does not exist too.

Here is a counterexample to his claim (in machine readable form, the degrees are high)

a=16384 * y^6 * x^6 * (x - y)^6 * (x + y)^6 * (x^2 + y^2)^6 * (x^4 - 10*x^2*y^2 + y^4)^3 * (x^8 - 8*x^6*y^2 + 30*x^4*y^4 - 8*x^2*y^6 + y^8)^3 * (x^8 + 4*x^6*y^2 + 54*x^4*y^4 + 4*x^2*y^6 + y^8)^3
b=(x^8 - 20*x^6*y^2 + 6*x^4*y^4 - 20*x^2*y^6 + y^8)^2 * (x^8 - 4*x^6*y^2 + 38*x^4*y^4 - 4*x^2*y^6 + y^8)^2 * (x^16 - 8*x^14*y^2 - 4*x^12*y^4 + 72*x^10*y^6 + 902*x^8*y^8 + 72*x^6*y^10 - 4*x^4*y^12 - 8*x^2*y^14 + y^16)^2 * (x^16 + 8*x^14*y^2 + 220*x^12*y^4 - 1480*x^10*y^6 + 3526*x^8*y^8 - 1480*x^6*y^10 + 220*x^4*y^12 + 8*x^2*y^14 + y^16)^2
c=a+b=(x^8 + 4*x^6*y^2 - 42*x^4*y^4 + 4*x^2*y^6 + y^8)^4 * (x^16 - 16*x^14*y^2 + 172*x^12*y^4 - 304*x^10*y^6 + 1318*x^8*y^8 - 304*x^6*y^10 + 172*x^4*y^12 - 16*x^2*y^14 + y^16)^4

Here we have deg(c)=96, deg(rad(abc))=98.

Answer 1

EDIT: I removed my wrong answer, I'm not even claiming anymore that the first case (where $xy$ divides $\text{rad}(abc)$) does not exist. Still, I believe that tricks like setting $y=\lambda x$ for a suitable complex number $\lambda$ and applying the univariate ABC inequality might help here. However, my original argument suffers if too high powers of $x$ divide $a(x,\lambda x)$ and $b(x,\lambda x)$.