Here is an exercise I think I managed to solve a part of it.
Let $A$ be a ring containing a field $K$ an $\alpha \in A$ any element, then the subring $K[\alpha] \subset A$ generated by $K$ and $\alpha$ is isomorphic to $K[T]$ or is a field extension of the form $K[T]/(f)$, where $f$ is irreducible.
What I managed to do so far: I divided the problem into two parts, where $\alpha$ is nilpotent or not. If it is not a nilpotent element, then it is clear to me that $K[\alpha]$ is isomorphic to $K[T]$, my problem is in the other case when $\alpha$ is nilpotent. I set $m$ to be the smallest natural number such that $\alpha^m=0$. Then I thought to choose $f=T^m$. However, then $f$ is not irreducible. I do not know how to overcome this obstacle.
Who says that $f$ has to be irreducible?
Every ring $K[\alpha]$ generated by a single element over a field is isomorphic to $K[T]/(f)$ where $f$ is the minimal polynomial of $\alpha$ (possibly $0$ if $\alpha$ is transcendental). So in particular it is not true that $K[\alpha] \cong K[T]$ when $\alpha$ is not nilpotent.
And, on the other hand, if $f$ is irreducible then $(f)$ is a prime ideal (since $K[T]$ is a UFD). If $(f)$ is prime then $K[T]/(f)$ is an integral domain and hence has no nilpotents. Thus if $\alpha$ is nilpotent, $K[\alpha]$ will never be isomorphic to $K[T]/(f)$ where $f$ is irreducible.