Let v be any nonzero element of S, which cannot be nilpotent as S is an integral domain. Therefore the ring S/v, with powers of v in the denominator, is a finitely generated R algebra. (Use the original generators and 1/v.)
Note that R embeds in S embeds in S/v. Normalize S/v over R, so that there is some nonzero element f in R, and some intermediate transcendental extension Rt between R and S/v, with S/v/f integral over Rt/f. (This can be done via the previous theorem.)
Let M be a maximal ideal in R missing f, and let K be the field R/M. Thus R maps onto K, and f maps to something nonzero in K. Again, review the previous theorem, and extend the map on R to a ring homomorphism g(S/v) into C, where C is the algebraic closure of K.
Since v is an invertible element in S/v it is a unit, the image of v is a unit, and v/1 is not in the kernel of g. Restrict g to S, and v still misses the kernel of g. If this kernel is a maximal ideal, then v is not in jac(S), and we're done.
Show the kernel of g is maximal by showing the image is a field. Remember that S is a finitely generated R algebra. Since g maps M to 0, the action of g only depends on the value of S mod M. In other words, g acts on S, when S is viewed as a finitely generated K algebra. No information is lost. Again, g carries S into C. An earlier theorem proves the image of S is a field, hence the kernel is a maximal ideal. The arbitrary element v is not in jac(S), and jac(S) = 0.