Let $R$ be a Noetherian ring. If $I$ is an ideal of $R$, then we know that for some integer $n>1$, $(\sqrt I)^n \subseteq I$.
My question is:
In a Noetherian ring, is it true that for every non-zero, proper ideal $I$, $\exists n>1$ such that for every $x \in \sqrt I \setminus I$, $x^n \in I$ but $x^{n-1} \notin I$ ?
On
No : take $R=\mathbb{Z}$, which is certainly noetherian, $I= (72)$, then $\sqrt{I} = (6)$.
Now if $n$ is such that the property you say holds, then $6^n \in I$, so $n\geq 3$. In particular, since $6^3 = 216 = 72\times 3$, your $n$ is $3$. But now with $x= 36$, $36\in \sqrt{I}\setminus I$, but $36^2 = 1296 = 72\times 18 \in I$.
Those numerical values aren't picked stupidly of course, I simply chose two distinct primes $p,q$ and picked $I=(p^3q^2)$. This way I knew that $n$ would have to be at least three for $p^3q^2$ to divide $(pq)^n$, but then $(p^2q)^2$ would be in $I$. And to have this idea I simply tried to prove the result you stated and saw where it went wrong.
Certainly not.
Even for something as simple as $\mathbb Z/8\mathbb Z$ this isn't true. Now both $2$ and $4$ are in $\sqrt{\{0\}}$ for this ring, and $2^3=0$ but $2^2\neq 0$, but $4^3=4^2=0$.
In fact, the only case in which it is going to be true is if $\sqrt{I}^2\subseteq I$. For, if there exists an $n>2$ and an element such that $x^n\in I$ but $x^{n-1}\notin I$, it must be true that $y=x^{n-1}$ satisfies $y^2\in I$, thwarting the condition.