Specifically, is it possible for a non-Noetherian ring $R$ to have $R[x]$ Noetherian? Every reference I've seen for the Hilbert basis theorem only states the direction "$R$ Noetherian $\Rightarrow$ $R[x]$ Noetherian", which would certainly seem to imply that the converse is false. Unfortunately, it's tough to think about non-Noetherian rings, and what I'm sure is most people's favorite example of one, $K[x_1,x_2,\ldots]$ for a field $K$, is obviously not going to help us here.
There are some sources which give the converse. See e.g. pp. 64-65 of
For that matter, see also pp. 32-33 of loc. cit. for the Chinese Remainder Theorem and its converse. (And I am not the only one to do this...)
Note that in both cases the converse is left as an exercise. I think (evidently) that this is the right way to go: it may not be so easy for the journeyman mathematician to come up with the statement of the converse, but having seen the statement it is a very valuable exercise to come up with the proof. (In particular, I believe that in a math text or course at the advanced undergraduate level and beyond, most exercises should indeed be things that one could find useful later on, and not just things which are challenging to prove but deservedly forgettable.)
Finally, by coincidence, just yesterday in my graduate course on local fields I got to the proof of the "tensor product theorem" on the classification of norms in a finite-dimensional field extension (which came up in a previous MO answer). The key idea of the proof -- which I found somewhat challenging to write; I certainly admit to the possibility of improvements in the exposition -- seems to be suspiciously close to the valuation-theoretic analogue of the converse of the Chinese Remainder Theorem! See pp. 18-19 of
if you're interested.