Let $f(x,y)\in \mathbb{C}[x,y]$ be an irreducible polynomial. Then $f$ corresponds to a curve in $\mathbb{A}^2$. For $(a,b)\in \mathbb{Z}^2$, consider $g = f(a+x,b+y)$ the translation of $f$ by $(a,b)$. Under what circumstances does $V(g) = V(f)$?
I can think of examples where this holds, for instance if $f=y-x$ then $(a,a)$ will work for any $a \in \mathbb{Z}$. However, I would like to conclude that lines with rational gradient is the only class of examples where this holds.
I think this boils down to looking at $f(x,y) = f(a+x,b+y)$, (looking for the intersection of the varieties) and showing that this cannot happen except when $a=b=0$. I think I have done this for when $f$ is a conic section but I'm struggling for higher degree polynomials. Does anyone have any suggestions, is this perhaps a standard problem in algebraic geometry that I have not found?
Yes, your impression is correct. This can only happen when $V(f)$ is a line with slope $b/a$.
Let $P=(x_0,y_0)$ be an arbitrary point on $V(f)$. Let $L$ be the line with slope $b/a$ through $P$. It follows that all the points $P_n=(x_0+na,y_0+nb)$, $n\in\Bbb{Z}$, are on $V(f)$ (induction on $|n|$). Therefore $V(f)$ and $L$ intersect at infinitely many point. This is in violation of Bezout's theorem stating that, unless $V(f)=L$, the intersection $V(f)\cap L$ has cardinality $\deg f$ (counted projectively and with multiplicities).
Observe that the characteristic of the ground field plays a role here. In characteristic $p$ we have counterexamples. For example, Artin-Schreier curves $C:y^p-y=f(x)$ have the property that $(x_0,y_0)\in C$ if and only $(x_0,y_0+1)\in C$.