Why don't we have an isomorphism between $R[x]$ and $ R[[x]]$? $R$ is a ring. Why don't we have an isomorphism 
$$
R[x] \cong R[[x]]\ ?
$$

The zero ring is the only commutative ring $R$ with $R[x] \cong R[[x]]$.
Proof: Assume $R[x] \cong R[[x]]$. It is known (Exercise 1.4 in Atiyah's book) that the Jacobson radical of $R[x]$ is equal to its nilradical. Thus, it would follow the same for $R[[x]]$. Since $x$ lies in the Jacobson radical of $R[[x]]$ (in fact, $1+xf$ is invertible for all $f \in R[[x]]$), it follows that $x$ is nilpotent, i.e. $x^n=0$ for some $n \geq 0$. This shows $R=0$.
The same proof shows: If $R,S$ are commutative rings with $R[x] \cong S[[x]]$, then $S=0$ and hence $R=0$.
