Although there is a good reason that $(x,y)^2$ has a smooth blow-up. It is a power of an ideal which itself has a smooth blowup. See for example Hartshorne, Algebraic Geometry, Chapter II, Section 7, Exercise 7.11.
I suspect that, on smooth surfaces, one can probably say more, via "Zariski-factorization" type ideas, but I'm not sure what the right answer would be.