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.