If $k$ is a field the ideals $I\subset A=k[X,Y]$ which are projective modules over $A$ are exactly the principal ideals.
Indeed every projective module over $A$ is free and a free ideal in a domain can only be of dimension $0$ or $1$, because two non-zero elements $f,g$ are always linearly dependent: $g.f+ (-f).g=0$ !
So the free (= projective) ideals in $A$ are $I=(0)$ with $dim_A(0)=0 $ and $I=fA, \; f\neq 0$ with $dim_A (fA)=1$ (where $\lbrace f \rbrace $ is a basis of $I$).
Since I'm in an aggressive mood [:-)], I'll use a sledge-hammer to justify my claim above that $I$ is free if it is projective: the Quillen-Suslin theorem, which indeed says that any finitely generated projective module over a polynomial ring $k[X_1,X_2,...,X_n]$ ($ k$ a field) is free.
I'm in a more peaceful mood now, and I'll give a more elementary argument that projective implies free.
If a non-zero ideal in a domain is projective, it must be projective of rank one (this is proved in Bourbaki, Algèbre Commutative,chap.II, §5, Théorème 4).
And now you just use that projective modules of rank one in a UFD are free (in geometric language: the Picard group of a UFD is trivial. You can find that in this book , on page 309).