It is undecidable.

The only integral point on $x^3+x=y^2$ is $(0,0)$.

Let $F(\vec{y})=0$ be undecidable diophantine equation with positive coefficients and not depending on $x$.

Take $f(x)=x^3+x$ and $g(\vec{y})=F^2$ leading to $x^3+x=F^2(\vec{y})$.

To get $F$ from $F'$ with negative coefficients use sum of squares replacing each negative coefficient $c_i$ with variable $v_i$ and add the square $(v_i + c_i)^2$.

It is undecidable.

The only integral point on $x^3+x=y^2$ is $(0,0)$.

Let $F(\vec{y})=0$ be undecidable diophantine equation with positive coefficients and not depending on $x$.

Take $f(x)=x^3+x$ and $g(\vec{y})=F^2$ leading to $x^3+x=F^2(\vec{y})$.

To get $F$ from $F'$ with negative coefficients use sum of squares replacing each negative coefficient $c_i$ with variable $v_i$ and add the square $(v_i + c_i)^2$.

Even simpler, the integral solutions of $x^2=1+d^2 y^2$ are $(\pm 1,0)$.