$R$ doesn't need to be connected, so long as $k$ is (and if $R$ is connected then $k$ is, since a nontrivial idempotent of $k$ would be a nontrivial central idempotent of $R$). Also, $R$ doesn't need to be finitely generated.
So assume $k$ is connected and not a field. Then $\operatorname{rad}k\neq0$ and so, since the Jacobson radical of an artinian ring is nilpotent, we can pick a nonzero nilpotent element $a\in k$. Let $n>1$ be the smallest integer such that $a^n=0$.
Now consider the short exact sequence$$0\to\operatorname{ann}_R(a)\to R\to aR\to0.$$
If $R$ is right hereditary then $aR$, as a right ideal of $R$, is projective, and so this short exact sequence splits (as a sequence of right $R$-modules).
So $R\cong\operatorname{ann}_R(a)\oplus aR$ as a right $R$-module.
But $\operatorname{ann}_R(a)$ is annihilated by $a$, and $aR$ is annihilated by $a^{n-1}$, so $R$ is annihilated by $a^{n-1}$, which is not true, since $a^{n-1}\neq0$.