A (two-sided) ideal $I$ of a ring with identity $R$ has the right Artin-Rees property if for any right-ideal $E$ of $R$, there exists an integer $n\geq1$ such that $E\cap I^n\subseteq EI$.
If $R$ is a right-Noetherian ring and the Jacobson radical $\text{Jac}(R)$ satisfies the right Artin-Rees property, then $\bigcap^{\infty}_{n=1}\text{Jac}(R)^n = 0$.
I've somehow proved this without the Noetherian property, so my proof most likely is wrong.
Let $J:=\bigcap^{\infty}_{n=1}\text{Jac}(R)^n$ and suppose $J\not= 0$. Let $x\in J\setminus 0$ and consider the right ideal $xR$. By assumption, we have that $xR \cap \text{Jac}(R)^k \subseteq xR\cdot\text{Jac}(R)$ for some $k$. But $xR\subseteq J\subseteq \text{Jac}(R)^k$ so we get that $xR\subseteq x\text{Jac}(R)$.
Therefore, since $1\in R$, we have that $x=xj$ for some $j\in\text{Jac}(R)$. As the Jacobson radical is right-quasi-regular, $1-j$ has a right inverse $y$. In particular, $x(1-j)=0$ and hence $x(1-j)y=x=0$, contradicting the assumption $x\not=0$.
Is this correct? What is missing?
I guess your teacher was thinking to Nakayama’s lemma. By assumption, there exists $n$ such that $$ J\cap (\operatorname{Jac}(R))^n\subseteq J\operatorname{Jac}(R) $$ On the other hand, $J=J\cap (\operatorname{Jac}(R))^n$ and $J\operatorname{Jac}(R)\subseteq J$, so we conclude $$ J\operatorname{Jac}(R)=J $$ Since $J$ is finitely generated, because $R$ is right Noetherian, we have $J=0$ by Nakayama’s lemma.
You showed instead that every cyclic submodule of $J$ is zero (using an abridged version of Nakayama’s lemma) and the proof is correct. So the hypothesis that $R$ is right Noetherian can be removed.