Let's consider the ring $R = \begin{bmatrix}\Bbb{Q} & 0\\\Bbb{Q} & \Bbb{Z}\end{bmatrix} = \left\{\begin{bmatrix}q & 0\\p & z\end{bmatrix} {\Big|\,} q,p \in \Bbb{Q}, z \in \Bbb{Z}\right\}$ and the right $R$-module $M = \begin{bmatrix}0 & 0\\\Bbb{Q} & \Bbb{Z}\end{bmatrix} = \left\{\begin{bmatrix}0 & 0\\p & z\end{bmatrix} {\Big|\,} p \in \Bbb{Q}, z \in \Bbb{Z}\right\}$.
I want to prove that $M$ is noetherian, that is, every ascending chain of submodules of $M$ in stationary. But how can I do it? There are infinite submodules of $M$.
By this solution you can conclude your ring $R$ is right Noetherian, and $M$, being a right submodule of $R$ obviously has to be right Noetherian as well.