Conjecture. Let $a,b, c\in \Bbb{Z}, b \neq 0$, The following conditions are equivalent:
(1) $d = \gcd(a,b)$ divides c.
(2) There's a polynomial in $f \in \Bbb{Z}[X,Y]$ with $c$ constant term, such that $f(a,b) = 0$.
(3) There's a polynomial $f \in \Bbb{Z}[X^{-1}, Y^{-1}, X, Y]$ with $c$ constant term such that $f(a, b) = 0$. Thanks @ajotatxe
(3) There's a power series in $f \in \Bbb{Z}[[X,Y]]$ with $c$ constant term such that $f(a,b) \to 0$.
Corollary. $\gcd(a,b) = 1$ if and only if there's a polynomial (resp. series) with constant term $1$ in either of the above two rings, that vanishes at $(a,b)$.
Motivation: this came up studying the fact that $p$ is the first prime in a pair of twin primes iff $p(p+2)X + (p-1)!Y = 1$ has an integer solution $X,Y$. Sub in $Z$ for $p$ and consider taking the the product $f(Z) = \prod_{n,m \in \Bbb{Z} \setminus 0} (1 - Z(Z+2)n -(Z-1)!m)$. Then $f$ is a power series in $X = Z(Z+2), \ Y = (Z-1)!$, and $f$ has infinitely many integer solutions if and only if there are infinitely many twin prime pairs.
Thanks.
$3$ is not equivalent to $1$. Let $$f=1+2X^{-1}-9Y^{-1}$$ and take $a=4$, $b=6$. We have $f(4,6)=0$ and the constant term of $c$ is $1$, but $\gcd(a,b)=2$.