When quotient of a $k$-algebra by any maximal ideal is $k$? Let $k$ be a valued field.
Is there a special term for a commutative (Banach) $k$-algebra $A$ such that for any maximal ideal $m$ we have $A/m=k$?
Is there an easy to check criterion that would imply this property?
 A: I don't think that there is a standard term for what you are looking for, but I would be inclined to call such an algebra "compact".  My reason is the following.  Suppose that $X$ is a manifold (resp. locally compact Hausdorff space).  Then the maximal spectrum of the ring $C^\infty(X)$ of smooth real-valued functions (resp. $C^0(X)$ of continuous real-valued functions) with the Zariski topology is homeomorphic to the Stone--Cech compactification $\beta X$ of $X$.  If $X$ is not astronomically large*, then the points $\mathfrak m \in \beta X$ for which the quotient field $C^\infty(X)/\mathfrak m$ (resp. $C^0(X)/\mathfrak m$) is isomorphic to $\mathbb R$ are precisely the points in $X \hookrightarrow \beta X$.  In particular, $X$ is compact iff $\beta X = X$ iff every maximal ideal comes from a point of $X$ iff every maximal ideal has quotient field $\mathbb R$.
I said this over $\mathbb R$, but really any infinite field $\mathbb K$ will do.  (Then I need to take "isomorphic" to mean "isomorphic as $\mathbb K$-algebras", so that if something is isomorphic to $\mathbb K$, then it is so uniquely.  Over $\mathbb R$, I can in fact take "isomorphic" to mean "isomorphic as rings", since $\mathbb R$ has no nontrivial ring automorphisms.)
*Footnote: The claim fails if there is a measurable cardinal $\lambda > |\mathbb K|$ with $|X| \geq \lambda$.  Note that in many universes there are no uncountable measurable cardinals at all, and if uncountable measurable cardinals exist, then they are strongly inaccessible, and hence very very large.
