These rings are called simple rings. As Torsten says in the comments, a nice class of examples which are not division rings is the matrix algebras $M_n(k)$. More generally, by the Artin-Wedderburn theorem the artinian simple rings are precisely the rings of the form $M_n(D)$ where $D$ is a division algebra. If the center of $D$ is $k$ then these are central simple algebras over $k$ and are classified by the Brauer group of $k$.
To prove that $M_n(D)$ is simple it's cleaner to prove a more general result:
Claim: The two-sided ideals of $M_n(R)$, for any ring $R$, are of the form $M_n(I)$ where $I$ is a two-sided ideal of $R$.
Corollary: $M_n(R)$ is simple iff $R$ is simple.
Proof. Let $X \in M_n(R)$ be any element. The ideal generated by $X$ consists of linear combinations of elements of the form
$$e_{ij} X e_{kl}$$
where $1 \le i, j, k, l \le n$; this matrix has only a single nonzero component, namely the $il$ entry, whose value is $X_{jk}$. So by picking $i, j, k, l$ appropriately we see that we can arrange for any particular component of $X$ to end up in any other component; in other words, the ideal generated by $X$ is $M_n(I)$ where $I$ is the ideal of $R$ generated by the components of $X$. The desired result follows upon taking sums of ideals, which gives more generally that the ideal generated by any collection of matrices is $M_n(I)$ where $I$ is the ideal of $R$ generated by their components. $\Box$
I said above that the artinian simple rings are the rings of the form $M_n(D)$, so let's close with a non-artinian example. Maybe the most famous non-artinian simple rings are the Weyl algebras, of which the one-variable version can be written
$$k[x, \partial]/(\partial x - x \partial = 1)$$
where $k$ is a field of characteristic zero; this can be thought of as the algebra of differential operators on $k[x]$, with $\partial$ acting by differentiation by $x$.
Claim: With the above hypotheses, the Weyl algebra is simple.
Proof. Let $f = \sum f_{ij} x^i \partial^j$ be an element of the Weyl algebra; we will show directly that if $f$ is nonzero then the ideal it generates is the entire Weyl algebra. First, observe that the monomials $x^i \partial^j$ form a basis of the Weyl algebra; there are various ways to prove this, it is a PBW-type result. So $f = 0$ iff $f_{ij} = 0$ for all $i, j$.
The defining relation of the Weyl algebra can be written $[\partial, x] = 1$, where $[a, b] = ab - ba$ is the commutator bracket. Now, the commutator bracket $[\partial, -]$ is always a derivation, so it follows that
$$[\partial, x^i] = \partial x^i - x^i \partial = ix^{i-1}$$
while $[\partial, \partial] = 0$ and hence $[\partial, \partial^j] = 0$. This gives
$$[\partial, x^i \partial^j] = ix^{i-1} \partial^j$$
and hence
$$[\partial, f] = \sum f_{ij} ix^{i-1} \partial^j.$$
That is, computing the commutator by $\partial$ has the effect of differentiating the polynomial part of $f$ (and note that $[\partial, f] = \partial f - f \partial$ lies in the ideal generated by $f$). So we can repeatedly apply $[\partial, -]$ to $f$ until all of its polynomial parts vanish except the ones of highest degree; hence we can assume WLOG that $f$ in fact has the form
$$f = \sum f_j \partial^j$$
where at least one $f_j$ is zero (this is where we need both the assumption that the $x^i \partial^j$ form a basis and that $k$ has characteristic zero). At this point we can now instead apply the derivation $[-, x]$, which satisfies $[\partial, x] = 1$ and hence by induction
$$[\partial^j, x] = j\partial^{j-1}$$
which gives
$$[f, x] = \sum f_j j \partial^{j-1}.$$
So we can again repeatedly "differentiate" until $f$ is a nonzero constant, which clearly generates the entire Weyl algebra. $\Box$
Best Answer
Well since $R$ has exactly two right ideals we know a priori that the two ideals are just $\{0\},R$ since these two are always ideals. Therefore if $I$ is a non zero ideal of $R$ then $I=R$.
Let $a\in R, \ a\neq0$. We will show that $a$ is invertible and therefore $R$ is a division ring. Consider the ideal $\langle a\rangle=\{ar:r\in R\}$. Since $\langle a\rangle\neq\{0\} \Longrightarrow \langle a\rangle=R \Longrightarrow 1\in \langle a\rangle\Longrightarrow 1=ar$ for some $r\in R.$ Now by considering the ideal $\langle r\rangle$ there is some $s\in R$ such that $rs=1$. It remains to show that $a=s$. Use that $1\cdot s=s$ and $a=a\cdot1$.