Not sure if this makes sense, but is it possible Fermat's Last Theorem to fail with a parametrization over some extension of $\mathbb{Z}$, i.e. are there not all constant $x(t),y(t),z(t) \in K[t]$ where $K$ is an extension of $\mathbb{Z}$ s.t. $$x(t)^p + y(t)^p=z(t)^p, x(t)y(t)z(t) \ne 0, p > 2$$

I suppose this would mean in an extension of $\mathbb{Q}$ the curve $x^p + y^p = z^p$ will be of genus $0$.

Tried equating coefficients with $p=3$, got relatively small undetermined system but failed to solve it or compute groebner basis. The system has solutions like $x(t)=0$.

Not sure if this makes sense, but is it possible Fermat's Last Theorem to fail with a parametrization over some extension of $\mathbb{Z}$, i.e. are there not all constant $x(t),y(t),z(t) \in K[t]$ where $K$ is an extension of $\mathbb{Z}$ s.t. $$x(t)^p + y(t)^p=z(t)^p, x(t)y(t)z(t) \ne 0, p > 2,\gcd(x(t),y(t),z(t))=1 $$

I suppose this would mean in an extension of $\mathbb{Q}$ the curve $x^p + y^p = z^p$ will be of genus $0$.

Tried equating coefficients with $p=3$, got relatively small undetermined system but failed to solve it or compute groebner basis. The system has solutions like $x(t)=0$.

EDIT: The coprimality condition is to avoid scaling a single solution by a polynomial.