This question is inspired by For $x,y\ge 2$ does $x^4+x^2y^2+y^4$ ever divide $x^4y^4+x^2y^2+1$?.
Let $f(x,y), g(x,y)\in \mathbb{Z}[x,y]$ with no common factor, such that every intersection point between the curves $f(x,y) = 0$ and $g(x,y) = 0$ has multiplicity one. Can we show that for every $\epsilon > 0$ there is a nonempty Zariski open subset $U_\epsilon$ and a constant $c_\epsilon$ such that for any integer point $(x,y)\in U_\epsilon(\mathbb{Z})$ we have $\gcd(f(x,y),g(x,y)) \le c_\epsilon \max(|x|,|y|)^{2+\epsilon}$?
A closely related question is
Let $S\subseteq \mathbb{Z}^2$ be a finite set of integer points. Can we show that for every $\epsilon > 0$ there is a nonempty Zariski open subset $U_\epsilon$ and a constant $c_\epsilon$ such that for any integer point $(x,y)\in U_\epsilon(\mathbb{Z})$ we have $\prod_{(a,b)\in S}\gcd(x-a,y-b) \le c_\epsilon \max(|x|,|y|)^{2+\epsilon}$?
The two questions are equivalent if the intersection of the curves $f(x,y) = 0, g(x,y) = 0$ is exactly the set of points $S$. I'll focus on the second question since it feels more intuitive to me.
Since these questions are likely to be hard, it would also be interesting to know if either is implied by a well-known conjecture.
Can we at least say anything about the special case $f(x,y) = x^3-x, g(x,y) = y^3-y$?
In this special case, there is a set of eight bad lines on which the gcd grows like $\max(|x|,|y|)^3$ and a set of eight bad hyperbolas on which the gcd grows like $\max(|x|,|y|)^{\frac{5}{2}}$. Aside from these bad cases, the worst point $(x,y)$ with $1 < x < y \le 500000$ that came up in a computer search was $(x,y) = (76941, 319301)$, with $\gcd(x^3-x,y^3-y) = 26793198284040 < 319301^{2.44}$.
Heuristic Argument
Suppose we try to construct a bad point $(x,y)$ as follows. Start by picking a collection of prime powers $p_i^{e_i}$ whose product will be $\prod_{(a,b)\in S}\gcd(x-a,y-b)$. Next, for each prime power, we choose one of the $|S|$ points to be congruent to $x,y$ modulo $p_i^{e_i}$. Finally, we solve for $x,y$ using the Chinese Remainder Theorem, and hope that $|x|,|y| \le \prod_i p_i^{\frac{e_i}{2+\epsilon}}$. If there are no algebraic conspiracies, we expect the total number of successes of this procedure to be something like
$$\prod_{p} \left(1+\sum_{e>0}\frac{|S|p^{\frac{2e}{2+\epsilon}}}{p^{2e}}\right) \approx \zeta\left(2-\frac{2}{2+\epsilon}\right)^{|S|} < \infty.$$
Actual Arguments
We can prove right away that the exponent $2$ can't possibly be improved if $S$ contains a set of four points, no three of which are collinear. Let $P_i = (a_i,b_i), i = 1, ..., 4$ be four such points in $S$, and let $Q = (c,d)$ be any integer point such that the conic passing through $P_1,P_2,P_3,P_4,Q$ is a hyperbola (any nonempty Zariski open set contains infinitely many such $Q$). By the theory of Pell equations, there are infinitely many integer points $(x,y)$ on this hyperbola. For such points, we can show (easy exercise) that up to constant factors we have
$$\gcd(x-a_i,y-b_i) \sim \max(|x|,|y|)^{\frac{1}{2}},$$
so the product of those four gcds grows like $\max(|x|,|y|)^2$.
We can also prove that the optimal exponent is, if not $2+\epsilon$, certainly at most $\sqrt{|S|}+\epsilon$. The main point is that if $p(x,y)$ is a nonzero polynomial and $k$ is an integer such that for each $(a,b) \in S$ the polynomial $p(x,y)$ is contained in the ideal $(x-a,y-b)^k$ (i.e., the point $(a,b)$ lies on the curve defined by $p(x,y) = 0$ with multiplicity at least $k$), then for any integer point $(x,y)$ we have
$$\gcd(x-a,y-b)^k = \gcd((x-a)^k,(y-b)^k,p(x,y)) \mid p(x,y),$$
so if $p$ has degree $d$ and $p(x,y) \ne 0$ then
$$\prod_{(a,b)\in S}\gcd(x-a,y-b) \ll |p(x,y)|^{\frac{1}{k}} \ll \max(|x|,|y|)^{\frac{d}{k}}.$$
In order to find such a polynomial $p$ of degree $d$, we need to solve a system of $|S|\frac{k(k+1)}{2}$ homogeneous linear equations in $\frac{(d+1)(d+2)}{2}$ variables, and we can do this if $d$ is large and $\frac{d}{k} \ge \sqrt{|S|}+\epsilon$.
