A ring with a unique nonzero proper ideal? Is there any commutative unital ring $R$ with a unique nonzero proper ideal? In particular, there must be a nonzero proper ideal, so fields don't work.
Clearly, such a ring must be local, and the unique (maximal) ideal $I$ contains every non-unit. If the ring is not reduced (i.e., there exists a nonzero nilpotent), then the nilradical is a nonzero proper ideal, so it must be equal to $I$. 
I know $R$ can't be an integral domain, since if $a\in R$ is a nonzero nonunit, then $(a^2)\neq (a)$ are distinct nonzero proper ideals. 
I'm not quite sure how to proceed from here. Can such an $R$ exist?
 A: $K[X]/(X^2)$, $K$ a field. It has only one prime ideal, generated by the class of $X$.
A: In the given setup , let $k=R/I$.
Following your argument $(a^2)\ne (a)$ for non-zero non-units $a$ implies $a^2=0$ for all non-units. (In particular, $R$ is reduced).
Pick $0\ne a\in I$. Then all elements of $I$ can be written as $ax$ with $x\in R^\times\cup\{0\}$ for if $x$ is a non-unit, the $x=ay\in I$ and $ax=a^2y=0$.
Define a multiplication on $I$ by $ax\odot ay=axy$ for $x,y\in R^\times \cup\{0\}$. This makes $(I,+,\odot,0,a)$  a field.
In fact, $(R,+,\cdot)\to (I,+,\odot)$, $x\mapsto ax$ is a ring homomorphism with kernel $I$, so the field $I$ is isomorphic to $R/I$.
Thus as rng, $R$ is an extension of $k$ by $k$:
$$0\to k\to R\to k\to 0$$


*

*The simplest case (i.e., where $R=k\oplus k$ holds for the additive groups) is $R=k[\epsilon]=k[X]/(X^2)$. (Bernard's example)

*For $k=\Bbb Z/p\Bbb Z$, we can easily write down $R=\Bbb Z/p^2\Bbb Z$. (For $p=2$, Nishant's example)

*For $k=\Bbb F_{q}=\Bbb F_p[X]/(f)$, we can wite down $R=(\Bbb Z/p^2\Bbb Z)[X]/(pf)$


I suppose more can be said if one has a closer look at the Ext functor ...
