What is true for a ring with exactly two right ideals

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 $a=a\cdot1$.