I found the following assertion at page 62 in Geometry of schemes by Eisenbud and Harris:
Let $R$ be a local $K$-algebra of vector-space dimension $2$, where $K$ is an algebraically closed field, and let $\mathfrak m$ be its maximal ideal. Then $R/\mathfrak m\cong K$ and $\mathfrak m^2=0.$
I can prove the first claim, by Nullstellensatz and the fact that $K$ is algebraically closed, but I failed to see why the square of $\mathfrak m$ is $0$. The book says that we might use Nakayama's lemma, but I could not perceive how to use that lemma.
So any hint or help will be mostly appreciated, and thanks in advance.