Does $R[[x]] \cong S[[x]]$ imply $R\cong S$ Let $R,S$ be commutative unitary rings. Is it true that

$$R[[x]] \cong S[[x]] \quad \Rightarrow \quad  R\cong S.$$

Here by $R[[x]], S[[x]]$ I mean the ring of formal power series and the isomorphisms as isomorphisms of rings.
In fact, in this question Does $R[x] \cong S[x]$ imply $R \cong S$? the answer for the polynomial ring is negative and I became curious about the formal power series case.
 A: This is not true. 
An explicit counterexample is given on p. 154 of the paper On power-invariance, by Eloise A. Hamann,     Pacific J. Math., Volume 61, Number 1 (1975), 153-159.
The example was given by Andy Magid at the   Commutative  Algebra   Conference  in  June   of  1974, based on Hochster's counterexample for polynomial rings.

More precisely, Hochster's counterexample provides (using hairy ball theorem) a projective non-free module $P$ over the commutative ring $A = \Bbb R[x,y,z]/(x^2+y^2+z^2-1)$ such that $A^3 \cong A \oplus P$ as $A$-modules.
Let $R$ be the complete symmetric algebra of the $A$-modules $P$, i.e. the completion  of  the symmetric  algebra $S_A(P)$ with  respect  to  the  ideal  generated  by $P$.
Let $S := A[[X,Y]]$.
Then, one can show that 
$$R[[T]] \cong A[[X,Y,Z]] \cong S[[T]]$$ as rings (not necessarily as topological rings), but 
$$R \not \cong A[[X,Y]] = S.$$

However, theorem 4 in Hamann's paper states that if $R$ has finitely many maximal ideals, then 
$R[\![x]\!] \cong S[\![x]\!] \implies R \cong S$, for every commutative ring $S$.
Moreover, theorem 1.4 in Joong Ho Kim's 1974 paper Power invariant rings shows that the same property holds if $R$ has a nilpotent Jacobson radical (i.e. the intersection $J$ of all maximal ideals of $R$ satisfies $J^n=0$ for some $n>0$). For instance, this holds if $R$ is artinian (e.g. a field)  or semiprimitive (e.g. $\Bbb Z$).
