Let f(x) = x^4-Mx+63.  f(x) can factor either as a product of two quadratic expressions or a linear times a cubic.  

Assume there is a linear factor.  Then for each integer factor of 63 there is a value M for which f(x) = 0. specifically if the factor is f then M = f^3 + 63/f.  The smallest positive value of M is 48, which occurs when f=3.

Assume two quadratic factors.  The linear terms must be additive inverses for the cubic coefficient of f(x) to be zero.  Let those values be a and -a.  Let f*g be a factorization of 63.  Then f(x) = (x^2-ax+f)*(x^2+ax+g).  The quadratic coefficient of f(x) is 0, which implies f+g-a^2 = 0.  Therefore for f(x) to have a factorization into two quadratic expressions there must be a factorization of 63 whose two terms sum to a square number.   There are in fact two ways: 1 and 63 or 7 and 9.  In the first case f(x) has a positive M of 64, and in the second case positive M is 16.

Then of all the ways to nontrivially factor f(x), the smallest value of M is 16, with the factorization (x^2+4x+7)*(x^2-4x+9).