Let $R$ be a right artinian algebra over an algebraically closed field $F$.
Claim R is algebraic over $F$ of bounded degree.
I am struggling with basics.
First of all, an elementary question:
Since $R$ is algebraic over $F$, and $F$ is algebraicly closed, does it trivially follow $R=F$?
Let me briefly mention my progress:
Let $\mathscr{J}(R)$ be the Jacobson Radical. $R/\mathscr{J}(R)$ is Jacobson semisimple. By Weddenburg-Artin; \begin{equation} R/\mathscr{J}(R) \cong \text{Mat}_{n_1}D_1\times\cdots\times\text{Mat}_{n_k}D_k \end{equation} for some division rings. Since $F$ is in center of $R$, it is contained in $D_1\times\cdots\times D_k$, center of $R/\mathscr{J}(R)$.
Question Is it possible to argue $F$ is contained in each $D_i$?
Question If $D_i$ is finite dimensional over $F$, then $D_i=F$. Is it finite dimensional?
Assuming $R/\mathscr{J}(R)\cong \text{Mat}_{n_1}F\times\cdots\times\text{Mat}_{n_k}F$, Cayley-Hamilton justifes existence of characteristic polynomial, hence for each $\bar{r} \in R/\mathscr{J}(R)$ there exists a polynomial of degree less than $n_1\cdots n_k$.
For right artinian ring, $\mathscr{J}(R)$ is nilpotent. Hence $s^n=0$ for all $s\in \mathscr{J}(R)$.
Question How can I lift the results to $R$ from $R/\mathscr{J}(R)$ and $\mathscr{J}(R)$?
First question: no - look at $M_n(F)$, for example. The key is that $M_n(F)$ has zero-divisors, so $(A - \mu_1)\cdots(A - \mu_r) = 0$ doesn't imply the existence of $i$ such that $A - \mu_i = 0$.
Second question: if $F \subset Z(A \times B)$ then $F(1, 0) \subset Z(A)$ and similarly $F(0, 1) \subset Z(B)$. (A slight correction to what you say in the question: $D_1 \times \cdots \times D_k$ isn't the centre of $R/\mathscr J(R)$, being in general noncommutative, but it does contain the centre).
Third question: it is; use the fact that $D_i$ is Artinian.
Last question: if $R/I$ and $I$ are both algebraic over $F$, that means that for any $r \in R$ there exists a polynomial $f_r$ over $F$ such that $f_r(r) \in I$, and for every $i \in I$ there exists a polynomial $g_i$ over $F$ such that $g_i(i) = 0$. Then if $r \in R$, it's immediate that $g_{f_r(r)}(f_r(r)) = 0$, and therefore $R$ is algebraic over $F$.