The question asks for necessary and sufficient conditions for a given curve to have integer points, that is, $(x_k,y_k)$ such that $x_k,y_k\in\mathbb{Z}$. For example, a necessary and sufficient condition for the curve $\mathcal{C}\colon y=ax^2+bx+c$ to have infinitely many integer points is precisely that it contain at least three integer points. Fewer than three is not enough, since for example $y=\frac{2x^2+1}{2}$, $y=\sqrt{2}x^2$, and $y=\sqrt{2}(x^2-1)$ are counterexamples to the claim.
I've also done the linear case.
However I would like to know what conditions a given cubic, or even a conic, must satisfy, e.g., what are necessary and sufficient conditions for a curve $f(x)=ax^3+bx^2+cx+d$ with $a\neq 0$? How many integer points are required so that we know the curve has infinitely many integer points?
For $y=ax^3+bx^2+cx+d$, four is enough, three is not.
The counterexamples for 3 are much as in the quadratic case. Let $p(x)$ be a cubic with integer coefficients and three integer roots, and consider $\sqrt2p(x)$.
If there are 4 then each integer point gives you a linear equation in the 4 unknowns $a,b,c,d$, with all coefficients integers, so all the unknowns must be rational (by Cramer's Rule). Putting over a common denominator, $T$, we have $y=(Ax^3+Bx^2+Cx+D)/T$, with $A,B,C,D,T$ all integers. If this quotient is an integer for $x=k$, then it is an integer for $x=k+nT$ for all integers $n$ (just plug in $x=k+nT$, multiply everything out, and see what happens --- or think about it in terms of congruences modulo $T$).
The same arguments work for polynomials of any degree.