A fellow grad student asked me this, I have been playing for a while but have not come up with anything. Note that $\mathbb{C}$ is homeomorphic to $\mathbb{C} – \{0\}$ in the Zariski topology – just take any bijection and the closed sets (finite sets) will biject as well. Concocting a similar thing for the plane is harder though.
I think I can show that the rational plane and the rational plane minus the origin are homeomorphic by enumerating the irreducible curves and using a back and forth argument, but I have not written it all up formally to see if I am missing something yet.
I know the question isn't natural from the point of view of algebraic geometry, because one of the objects isn't even a variety. I think it is still interesting just to see how weird the zariski topology really is.
Best Answer
This question was analyzed by Roger Wiegand in the context of algebraic surfaces over fields $k$ which are algebraic closures of finite fields.
[1] R. Wiegand, Homeomorphisms of affine surfaces over a finite field, J. London Math. Soc. (2) 18 (1978), no. 1, 28–32.
and in the followup paper:
[2] R. Wiegand, W. Krauter, Projective surfaces over a finite field. Proc. Amer. Math. Soc. 83 (1981), no. 2, 233–237.
In [1] he proves that the affine plane $A^2_k$ (over $k$) is Zariski-homeomorphic to any open nonempty subset of $A^2_k$ (Corollary 7). In particular, $A^2_k$ is homeomorphic to $A^2_k -\{(0,0)\}$. In particular, Greg's argument (as written) cannot be complete (as zero characteristic is not used anywhere in his sketch).
After proving the corollary Wiegand asks if the same holds for $k={\mathbb R}$ and ${\mathbb C}$.
In [2] it is proven (amon other things) that every proper nonempty open subset of $P^2_k$ is homeomorphic to either $P^2_k - \{point\}$, or to the affine plane $A^2_k$.