Does $R[x] \cong S[x]$ imply $R \cong S$?

Here is a counterexample.

Let $R=\dfrac{\mathbb{C}[x,y,z]}{\big(xy - (1 - z^2)\big)}$, $S=\dfrac{\mathbb{C}[x,y,z]}{\big(x^2y - (1 - z^2)\big)}$. Then, $R$ is not isomorphic to $S$ but, $R[T]\cong S[T]$.

In many variables, this is called the Zariski problem or cancellation of indeterminates and is largely open. Here is a discussion by Hochster (problem 3).

I found this paper by Brewer and Rutter that discusses related matters. They cite a forthcoming paper by Hochster which proves there are non-isomorphic commutative integral domains $R$ and $S$ with $R[x]\cong S[x]$.

Added

Hochster's paper is M. Hochster, Nonuniqueness of coefficient rings in a polynomial ring, Proc. Amer. Math. Soc. 34 (1972), 81-82, and is freely available.

There's been much work on this problem since the mentioned seminal work in the early seventies. Searching on the buzzword "stably equivalent" should help locate most of it. Below is a helpful introduction from Jon L Johnson: Cancellation and Prime Spectra