Consider the identity

$$(x+1)^5+(y+1)^5 = (2x + 6y) ( -x +y )^4 + f(x,y)$$ where $f(x,y)=-x^5 + 2*x^4*y + 12*x^3*y^2 - 28*x^2*y^3 + 22*x*y^4 - 5*y^5 + 5*x^4 + 5*y^4 + 10*x^3 + 10*y^3 + 10*x^2 + 10*y^2 + 5*x + 5*y + 2$.

The curve $C : f(x,y)=0$ is genus 0 and has infinitely many rational points. I don't know if it has infinitely many integral points: suppose not. For integral points on $C$ the identity is:

$$(x+1)^5+(y+1)^5 = (2x + 6y) ( -x +y )^4. \qquad (1)$$

Infinitely many solutions on $C$ with $\gcd(x+1,y+1)=1$ contradict $abc$, so $abc$ implies either finitely many integral solutions, or sufficiently large $\gcd(x+1,y+1)$ (clearing a small gcd will still give abc triples of sufficiently good quality).

The integral points might be growing exponentially, so abc for polynomials doesn't appear to apply.

Q1 Is there a finite extension of $\mathbb{Z}$ where $C$ has infinitely many integral points? How do I find integral points there?

Q2 Is there a similar identity where $f$ is divisible by a quadratic with infinitely many integral points?

I tried to solve Q2 by equating coefficients, but couldn't solve the system.

Other similar identities exist.

If $f$ is divisible by $g= x^2+xy -y^2+1$, integral points on $g$ are consecutive fibonacci numbers $(F_{2n},F_{2n+1})$. If the lhs is still a sum of powers of linear polynomials, it appears unlikely to me that the common factor will always be large, so probably such an identity doesn't exist.