As Toly commented, an infinite field extension of $k$ gives a counterexample. However, this is the only way to get a counterexample: if you assume all the residue fields of $A$ are finite over $k$, then $A$ must be finite-dimensional. Indeed, 0-dimensionality implies that $\text{spec }A$ is Hausdorff, and any Noetherian Hausdorff space is automatically finite and discrete. So $A$ is a finite product of Artinian local rings, each of which is finite-dimensional over its residue field.

(Proof that Noetherian and Hausdorff implies finite and discrete: fix a point $x$; by Hausdorffness, for any $y\neq x$, there is a closed neighborhood $C_y$ of $x$ that does not contain $y$. By Noetherianness, the intersection of all the $C_y$'s can be obtained by intersecting only finitely many of them, and is hence still a neighborhood of $x$. But this intersection is just $\{x\}$. This shows the space is discrete; finiteness then follows by Noetherianness.)