Here's how I would do it: we need to show every $\alpha \in R \setminus k$ has a multiplicative inverse, that is,
$\exists \beta \in R \mid \beta \alpha = \alpha \beta = 1; \tag 1$
we note in passing that we must have $\beta \in R \setminus k$; otherwise, $\alpha = \beta^{-1} \in k$. In any event, the hypothesis that $R$ is finite-dimensional over $k$ implies that the sequence of elements of $R$
$\{\alpha^i \mid i \in \Bbb N \} = \{ 1, \alpha, \alpha^2, \ldots, \alpha^m, \ldots, \} \tag 2$
must at some point become linearly dependent; that is, for some $n \in \Bbb N$ there are $p_0, p_1, p_2, \ldots, p_n \in k$ with
$\sum_0^n p_i \alpha^i = 0; \tag 3$
we set
$k[x] \ni p(x) = \sum_0^n p_i x^i; \tag 4$
then
$p(\alpha) = 0. \tag 5$
We can in fact take $\deg p(x)$ to be minimal; that is, we can assume that there is no $q(x) \in k[x]$, $\deg q(x) < \deg p(x)$ and $q(\alpha) = 0$. We also have $n = \deg p(x) \ge 2$, lest $\alpha \in k$.
I claim that for such $p(x)$, $p_0 \ne 0$; otherwise, we have $p_1(x) \in k[x]$ with $p(x) = xp_1(x)$; indeed,
$p_1(x) = \sum_1^n p_i x^{i - 1}; \tag 6$
then
$\alpha p_1(\alpha) = p(\alpha) = 0, \tag 7$
and since $0 \ne \alpha \in R \setminus k$, and $R$ is a domain, we conclude that
$p_1(\alpha) = 0, \tag 8$
which contradicts our assumption that $p(x)$ is of minimal degree. So, since now we have $p_0 \ne 0$, and
$p(x) = xp_1(x) + p_0, \tag 9$
we see that
$\alpha p_1(\alpha) = -p_0, \tag{10}$
whence
$\alpha(-p_0^{-1}p_1(\alpha)) = (-p_0^{-1}p_1(\alpha)) \alpha = 1; \tag{11}$
thus
$\alpha^{-1} = -p_0^{-1}p_1(\alpha),\tag{12}$
and $R$ is a division ring.
NB: Note I've used the fact that $R$ is a finite-dimensional vector space over $k$ (see ca. (2)-(3)); I don't see how this can be avoided. End of Note.