Suppose $P$ is a (two-sided) prime ideal of our Artinian ring, $R$. Choose any $x \in R - P$.
Claim 1: $Rx^n + P$ is a left ideal of $R$ for any $n \in \Bbb Z^+$. To see this, suppose we have:
$a = rx^n + p$
$b = r'x^n + p'$ for $r,r' \in R, p,p' \in P$.
Then $a - b = rx^n + p - r'x^n - p = (r-r')x^n + (p - p') \in Rx^n + P$, so this is an additive subgroup.
Also, for any $r'' \in R$:
$r''a = r''(rx^n + p) = (r''r)x^n + r''p \in Rx^n + P$
Now we have the descending chain:
$Rx + P \supseteq Rx^2 + P \supseteq \cdots \supseteq Rx^n + P \supseteq Rx^{n+1} + P \supseteq \cdots$
and since $R$ is Artinian it is left-Artinian, so this stabilizes for some positive integer $n$.
Hence $x^n = rx^{n+1} + p$ for some $r \in R, p \in P$, which means that:
$x^n - rx^{n+1} = (1 - rx)x^n \in P$.
$x^n \in P \implies x \in P$ so we must have: $1 - rx \in P$, by our choice of $x$.
Hence, $1 = rx + (1 - rx) \in Rx + P$ which shows $P$ is a maximal left ideal.
A similar proof using $xR + P$ shows $P$ is a maximal right ideal as well.
Since our ring $R$ is also postulated to be prime, $\{0\}$ is a prime ideal of $R$, and thus (by the above) $\{0\}$ is a maximal ideal, that is: $R$ is simple.
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(This proof is probably somewhat "inelegant", Artinian conditions aren't my strong suit).
