Suppose M is maximal in the ring of n indeterminants, yet M contracts back to 0 in Z. Let M1, M2, M3, … be the contraction of M into the subrings spanned by one indeterminant, then two indeterminants, then three, and so on. Thus Mn = M, and M0 = 0. Let Q0 be any prime ideal in Z, such as the multiples of p. Then let Q1 be a prime ideal in Z[x1], lying over Q0, and containing M1. Why can we do this? It is the going up property in a rather complicated theorem about chains of prime ideals in polynomial rings over a ufd. If you accept this result, then do it again, creating Q2 lying over Q1 and containing M2. This continues all the way up to Mn, and creates a larger ideal,which is a contradiction. Therefore the Maximal ideal M contains the multiples of some prime p in Z.
Mod out by the ideal generated by p, and M becomes a maximal ideal in Zp[x1,x2,…xn]. Let's try to generate the maximal ideal of the polynomial ring over Zp. We can always include p to produce the maximal ideal in the integer polynomials.
I'll do this for x and y, you can carry on from there. The contraction of M back into Zp[x] yields a prime ideal, and Zp[x] is a pid, so we're talking about 0, which is impossible since the contraction of M back to Zp is all of Zp, or the ideal generated by some irreducible polynomial g(x). So our finite field of order p is expanded to a larger finite field according to the degree of g.
You can find such a polynomial for each indeterminant, by contracting to that particular subring, but this doesn't necessarily span a maximal ideal. Consider x2+1 and y2+4 mod 7. Mod out by the first and get the finite field of order 49; but then the second splits into y+2x times y-2x. Seen another way, the ideal spanned by x2+1 and y2+4 can be enhanced by y-2x, and still does not span 1. So bring in y-2x instead, and then y2+4 is implied.
After x, y satisfies some irreducible polynomial h(y) over our extended finite field. If h has degree > 1, h extends the field further. This continues through all the indeterminants.