I'm trying to prove that a matrix ring of the form $\begin{pmatrix} R & R\\ 0 & Q \end{pmatrix}$ where $R/Q$ is an infinite degree field extension, is left Noetherian, but not right Noetherian. I was first looking at a concrete example, e.g. $\begin{pmatrix} \mathbb{R} & \mathbb{R}\\ 0 & \mathbb{Q} \end{pmatrix}$ which is also discussed here.
I think I have a solution for one part of the problem: If the extension $ R/Q$is of infinite degree, it means that there is an infinite sequence of field extensions $Q<K_1<K_2<…<R$. Looking at these $K_i$-s as $Q$-vector subspaces of $R$, using the construction given in the linked example, we get an infinite chain of right ideals, which means that it cannot be right Noetherian.
UPDATE: I think what I don't see right now is why such rings are left Noetherian. I think it would be convenient to tackle this using the fact that a module $M$ is (left) Noetherian iff for a submodule $N\leq M$, both $N$ and $M/N$ are (left) Noetharian.