Let $R = 2\mathbb{Z}$. Then $R[x]$ is not a noetherian ring.
I do not understand why this is so, because Hilbert's basis theorem says: If R Noetherian ring, then R[X] a is Noetherian ring (from wiki).
I suppose that $2\mathbb{Z}$ is principal ideal ring:
Let (2), (4), (6), ... are the ideals, therefore all elements are generated by one ideal, so $2\mathbb{Z}$ is principal ideal ring. And we conclude that $2\mathbb{Z}$
is a noetherian ring. Why can't use Hilbert's basis theorem for $R[x]$?
The Hilbert basis theorem only applies to unital rings (this is often not stated explicitly since unital is often included in the definition of "ring"). Since $2\mathbb{Z}$ is not unital, the Hilbert basis theorem does not apply in this case.
An example of a non-finitely generated ideal in $2\mathbb{Z}[x]$ is the entire ring $2\mathbb{Z}[x]$ itself. (Note that for this to be true, $2\mathbb{Z}[x]$ must be defined as the set of polynomials all of whose coefficients are in $2\mathbb{Z}$, rather than the ring obtained from $2\mathbb{Z}$ by freely adjoining a central element $x$. This distinction does not make a difference for unital rings.)