Let $R$ be a commutative Noetherian ring (with unity), and let $I$ be an ideal of $R$ such that $R/I \cong R$. Then is $I=(0)$?

Assume that $R/I$ and $R$ are isomorphic. Let us denote the isomophism by $f:R/I \to R$.

Let $\pi:R \to R/I$ denote the usual map $x \mapsto x + I$. This is a of course a surjective ring homomorphism.

The composition $f \circ \pi : R \to R$ is thus a surjective ring endomorphism (the composition of surjections is a surjection).

By the result quoted in the question $f \circ \pi$ is an isomorphism, in particular it is injective. It follows that $\pi$ is injective, otherwise the composition could not be injective.

The kernel of $\pi$ is thus $\{0\}$; it is also is $I$. Thus $I = \{0\}$