In the book Quasi Frobenius Rinngs- Nicholson and Yousif, Example 2.5 gives a ring as follows:
Let F be a field and assume that $a→ \bar{a}$ is an isomorphism $F → \bar{F} ⊆ F$, where the subfield $\bar{F}\neq F$. Let R denote the left vector space on basis {1, t}, and make R into an F-algebra by defining $t^2 = 0$ and $ta = \bar{a}t$ for all $a\in F$.
Is $R$ an Artin algebra over $F$? I'm confused because right and left $F-$module structures of $R$ seems different. On the other hand, for an $F$-algebra A, there exits a ring homomorphism from $F$ into center of $A$. So $xy=yx$ for all $x\in F$ and $y\in A$. For this example this is not true.
Although the dimension of $R$ over $F$ could be different on the right and left, is $R$ an Artin algebra over $F$ when $F_{\bar{F}}$ is finite dimensional?
I agree with what you have said: it is not really an $F$-algebra because the action of $F$ is not central on $R$.
Of course, one can try to use a nonstandard definition, but there is no hint of that, so a reader can only assume the most likely thing.
Whether it is Artinian on both sides or not depends on $[F:\bar F]$. If it is a finite extension, it is Artinian on both sides. If the extension is of infinite degree, then it is not right Artinian, in particular I think its right socle is an infinite direct sum of submodules.