In general, if $R$ is a noetherian ring and $\mathfrak m$ a maximal ideal of $R$, then $\hat R$ (the $\mathfrak m$-adic completion of $R$) is a noetherian local ring.
Write $\mathfrak m=(a_1,\dots,a_n)$. Then $\hat R\simeq R[[X_1,\dots,X_n]]/(X_1-a_1,\dots,X_n-a_n)$. A maximal ideal of $R[[X_1,\dots,X_n]]$ has the form $M=\mathfrak n R[[X_1,\dots,X_n]]+(X_1,\dots,X_n)$ with $\mathfrak n\subset R$ a maximal ideal. Since $(X_1-a_1,\dots,X_n-a_n)\subseteq M$ we must have $a_i\in\mathfrak n$ for all $i$, that is, $\mathfrak m=\mathfrak n$. As a consequence we get that $M=\mathfrak m R[[X_1,\dots,X_n]]+(X_1,\dots,X_n)$ and this is the only maximal ideal of $R[[X_1,\dots,X_n]]$ containing $(X_1-a_1,\dots,X_n-a_n)$.
Added in proof. I've found here on page $6$ a more general result: the $I$-adic completion $\hat R$ is quasi-local iff $R/I$ is quasi-local. (Quasi-local means local, but not necessarily noetherian.) In the noetherian case the proof goes exactly as I did before.
I believe, Lang meant that one can consider only the case of a local ring and the kernel being $\mathfrak{m}$ (i.e. $A:=A_{\mathfrak{p}}$). For the kernel: let $R$ be a ring of germs of infintely-differentiable functions, and take a homomorphism of $R$ to formal power series. That is embeddeble into the field of Laurent series, but the kernel is obviously not maximal ideal.
Best Answer
The $K$-vector space $R$ admits a filtration $R\supseteq \mathfrak{m}\supseteq \mathfrak{m}^2\supseteq\cdots \supseteq\{0\}$ where each quotient (for $i\geq 0$) $\mathfrak{m}^i/\mathfrak{m}^{i+1}$ is an $R/\mathfrak{m}\cong K$-vector space.
The fact that $R$ has dimension $2$ as a $K$-vector space restricts the length of this chain. Indeed, if $\mathfrak{m}^2\subsetneq \mathfrak{m}$, then we're basically done. (Why?) In order to rule out the case $\mathfrak{m}^2=\mathfrak{m}$, use Nakayama's lemma.
I hope this helps! (I mentioned the filtration because it's a very useful thing to keep in mind.)
Exercise 1: Let $R$ be a local Artinian ring with unique maximal ideal $\mathfrak{m}$. Prove that $\mathfrak{m}$ is a nilpotent ideal, that is, $\mathfrak{m}^i=0$ for some positive integer $i$. (You can assume that $R$ is also Noetherian if you are not aware that this is a Corollary of the hypothesis that $R$ is Artinian.)