Suppose there's another right ideal $J'$ such that $J' \not \subseteq J$. Take $x \in J'$ such that $x \notin J$. Then $x$ is left invertible, so there exists $r \in R$ such that $rx =1$. But $xr \neq 1$. Thus both $xr $ and $1 -xr$ are not left invertible (by your lemma), thus $xr, 1 - xr \in J$, so $1 \in J$, contradiction. Therefore $J$ is the unique maximal right ideal.
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 $R$.
Best Answer
The ring of $2×2$ matrices over reals is simple so it has one maximal two sided ideal, 0, but it has 2 maximal left (right) ideals.