Nope.

Suppose this were possible. For z \in Z, let R_{z} be the colors that appear in the row {(x,z) : x≠z \in Z}, and let C_{z} be the colors that appear in the column {(z,y) : y≠z \in Z}. The coloring condition is that each C_{z} and R_{z} is disjoint. Since there are finitely many possibilities for each set, find distinct z and z' where R_{z} = R_{z'}, and C_{z} = C_{z'}. What color is (z,z')? Well, it's in R_{z'} (and not C_{z'}), and it's in C_{z} (but not R_{z}). So it's both in R_{z} and not: a contradiction.

(Ninja'd. Drat. Well, different proof at least.)