Charlie, Thanks for the clarification.  I think I would have a solution (perhaps not unique) if I changed my conversion of 64 (base 10) from 40 (base 16) to 22 (base 31), keeping the other four I had from bases 11,12,13,14).

If I am correct that there are five squares (16,36,49,64,100) then the lowest possible set of two-digit bases (11..15) would be 65, and the lowest "available" square would then be 81 (which is 11+12+13+14+31).