This question is intimately tied to the idea of dimension. Given a commutative ring $R$, its (Krull) dimension is defined as the supremum of the lengths of chains of all prime ideals.
The name dimension for this quantity makes sense because the coordinate ring of an $n$-dimensional affine variety has Krull dimension $n$. Localization has a geometric interpretation, too: a prime ideal $\mathfrak{p}$ corresponds to an irreducible variety $V$. The localization $R_\mathfrak{p}$ consists of all rational functions that are defined at all points of $V$.
If $R$ is a local ring and its unique maximal ideal $\mathfrak{m}$ is the only prime ideal, then $R$ has dimension $0$, which, if $R$ is Noetherian, is equivalent to being Artinian.
The geometric interpretation above indicates how to produce a local ring with any number of prime ideals. Given a positive integer $n$, consider $n$-dimensional affine space $\mathbb{A}^n$ over a field $k$ of characteristic zero. Its coordinate ring is $k[x_1, \ldots, x_n]$, and if we localize at the prime (maximal, actually) ideal $(x_1, \ldots, x_n)$, we obtain the local ring $k[x_1, \ldots, x_n]_{(x_1, \ldots, x_n)}$ which consists of all rational functions on $\mathbb{A}^n$ that are defined at the origin. By the correspondence theorem for prime ideals in a localization, this ring has a chain of prime ideals
$$
(0) \subsetneq (x_1) \subsetneq (x_1, x_2) \subsetneq \ldots \subsetneq (x_1, x_2, \ldots, x_n)
$$
of length $n$.
As for an example where $R$ is not a domain: an affine variety is irreducible iff its coordinate ring is a domain, so we need to consider reducible varieties. Let $A = \frac{k[x,y]}{(xy)}$ which is the coordinate ring of the union of the lines $x = 0$ and $y = 0$ in the plane. Let $\mathfrak{m} = (x,y)$ and let
$$
R = A_\mathfrak{m} = \left(\frac{k[x,y]}{(xy)}\right)_{(x,y)}
$$
which is the local ring at the origin. Then $(x), (y)$, and $(x,y)$ are all prime ideals of $R$.
This group ring isn't commutative, so nilpotent elements aren't necessarily in the Jacobson radical, right?
While it does not necessarily contain all nilpotent elements, the Jacobson radical does contain each nil ideal. So, the nilpotent augmentation ideal is contained in the Jacobson radical.
Conversely, the augmentation ideal of the group ring over a field is maximal right ideal, so the Jacobson radical is contained in it. Thus the Jacobson radical is a maximal right ideal, and the ring has exactly one maximal ideal.
Best Answer
$S^{-1}R$ is the localization of $R$ at the prime ideal $p\mathbb{Z}$. Since $\mathbb{Z}$ is an integral domain, we can describe $S^{-1}R$ as a subring of its field of fractions, i.e. $\mathbb{Q}$. In this case $S^{-1}R$ consists of all those rational numbers which can be written as $\frac{a}{s}$ where $a\in R,s\in S$ that is $a$ and $s$ are integers with $s$ not divisible by $p$.
Now for the actual question regarding the maximal ideals:
Show that the units in $S^{-1}R$ are precisely those elements in $S^{-1}R\setminus pS^{-1}R$. This then shows that $pS^{-1}R$ is the unique maximal ideal of $S^{-1}R$ (why?).