Let $(R,m)$ be a local Noetherian integral domain. Let $I$ be an ideal in $R$, so $I \subseteq m$.
Then $R/I$ is also a local Noetherian ring, see (and the image of any surjective ring homomorphism of a Noetherian ring is Noetherian). But in order to be an integral domain, $I$ must be a prime ideal of $R$, and this is not assumed.
Then unique maximal ideal of $R/I$ is $m/I$.
Also recall that a zero-dimensional Noetherian ring is Artinian, Lemma 10.60.4., so $R/I$ is artinian.
Assume that the Krull dimension of $R/I$ is zero, $\dim(R/I)=0$. By the result mentioned (and proved) here, there exists $n \geq 1$ such that $(m/I)^n=\bar{0}$. This means that there exists $n \geq 1$ such that $m^n \subseteq I$.
Question: $m \supseteq I$, but powers of $m$ may not contain $I$, so we are not able to talk about a decreasing chain of ideals $(m/I) \supseteq (m/I)^2 \supseteq (m/I)^3 \cdots$? So perhaps if, for example, $m^2$ does not contain $I$, then we have $m/I=\bar{0}$ (namely, $m=I$)?
Thank you very much!
You have to be careful about how you define $(m/I)^n$. This does not correspond to $m^n$.
Now by definition
$$ (m/I)^n = \{\sum a_1 \cdot ... \cdot a_n | a_1, \dots , a_n \in m/I\}$$
To see what ideal of $R$ this corresponds to, just take the preimage and we see that the ideal in $R$ is going to be $I + m^n$ (since the ideal in $R/I$ contains zero, the preimage contains all of $I$).