The statement of the theorem doesn't seem to hint at any "basis" whatsoever, so why the "basis"?
The theorem:
If $A$ is a Noetherian ring, then $A[x]$ is also Noetherian.
Why is it called Hilbert’s “basis” theorem?
abstract-algebracommutative-algebranoetheriansoft-questionterminology
Best Answer
In this context a "basis" for an ideal $I$ is a generating set. Indeed, if our ideal is $\langle x,y,z \rangle$ (say), then we know every element of our ideal can be written as $ax+by+cz$ for some ring elements $a,b,c$. Keep in mind that unlike the vector space case, there might be multiple ways to do this!
So then with this terminology in hand, we come to some natural questions: Which ideals of which rings actually have a (finite) basis? Moreover, are there some bases that are "nicer" than others? And if so, how can we find the nice ones? All of these are incredibly important for actually doing computations with ideals and rings. So anytime you want to solve a system of polynomials (a very common problem!) you want to know how to find a nice basis for the ideal corresponding to that system, provided one exists.
This was viewed as quite a difficult problem, but Hilbert came and showed that the qualitative solution is easy! Provided every ideal of the ring $R$ we start with has a basis (that is, provided $R$ is noetherian) then every ideal of $R[x]$ also has a basis! The question of a "nice basis" was eventually solved as well, and these are called Gröbner Bases. We know how to find them in general, and they really do make computations easier! See, for instance, the excellent book Using Algebraic Geometry by Cox, Little, and O'Shea.
I hope this helps ^_^