Well, this is the opposite of what you asked, but there is an easy reduction in the other direction. Namely, if the result is true for countable fields, then it is true for all fields. I can give two totally different proofs of this, both very soft, using elementary methods from logic. While we wait for a solution in the requested direction, let me describe these two proofs.

**Proof 1**. Suppose k is any algebraically closed field, and J is an ideal in the polynomial ring k[x_{1},...,x_{n}]. Consider the structure (k[x_{1},...,x_{n}],k,J,+,.), which is the polynomial ring k[x_{1},...,x_{n}], together with a predicate for the field k and for the ideal J. By the downward Loweheim-Skolem theorem, there is a countable elementary substructure, which must have the form (F[x_{1},...,x_{n}],F,I,+,.), where F is a countable subfield of k, and I is a proper ideal in F[x_{1},...,x_{n}]. The "elementarity" part means that any statement expressible in this language that is true in the subring is also true in the original structure. In particular, I is a proper ideal in F[x_{1},...,x_{n}] and F is algebraically closed. Thus, by assumption, there is a_{1},...,a_{n} in F making all polynomials in I zero simultaneously. This is a fact about a_{1},...,a_{n} that is expressible in the smaller structure, and so it is also true in the upper structure. That is, every polynomial in J is zero at a_{1},...,a_{n}, as desired.

**Proof 2**. The second proof is much quicker, for it falls right out of simple considerations in set theory. Suppose that we can prove (in ZFC) that the theorem holds for countable fields. Now, suppose that k is any field and that J is a proper ideal in the ring k[x_{1},...,x_{n}]. If V is the set-theoretic universe, let V[G] be a forcing extension where k has become countable. (It is a remarkable fact about forcing that any set at all can become countable in a forcing extension.) We may consider k and k[x_{1},...,x_{n}] and J inside the forcing extension V[G]. Moving to the forcing extension does not affect any of our assumptions about k or k[x_{1},...,x_{n}] or J, except that now, in the forcing extension, k has become countable. Thus, by our assumption, there is a_{1},...,a_{n} in k^{n} making all polynomials in J zero. This fact was true in V[G], but since the elements of k and J are the same in V and V[G], and the evaluations of polynonmials is the same, it follows that this same solution works back in V. So the theorem is true for k in V, as desired.

But I know, it was the wrong reduction, since I am reducing from the uncountable to the countable, instead of from the countable to the uncountable, as you requested...

Nevertheless, I suppose that both of these arguments could be considered as alternative very soft short proofs of the uncountable case (assuming one has a proof of the countable case).