If you want us to be sure that we are correctly interpreting terminology that you have encountered "in the literature", you had better identify the piece of literature in question: no one can be responsible for knowledgeable about all possible ways that a term has been used across all mathematical papers and books.
But based on the context it is quite plausible that a zero dimensional ideal $I$ in a commutative ring $R$ means that the quotient ring $R/I$ has Krull dimension zero: prime ideals are maximal. At least this definition renders true the statement
$k[x_1,\ldots,x_n]/I$ has Krull dimension zero $\iff$ it is finite-dimensional as a $k$-vector space.
This is true over any field $k$. Recall that a ring is Artinian iff it is Noetherian and dimension zero. The implication $\Longleftarrow$ is then immediate. For $\Longrightarrow$ use the fact that an Artinian ring is a finite product of local Artinian rings to reduce to the case in which $I$ is a power of a maximal ideal (c.f. Theorem 8.35 here) and then to the case in which $I$ is maximal. In this latter case the result is Zariski's Lemma: see e.g. $\S$ 11.1 of loc. cit..
Also let me note that when $k$ is algebraically closed, the given conditions are also equivalent to: the corresponding affine algebraic set $V(I)$ is finite. This equivalence does not hold over any non-algebraically closed field. A simple example is $I = \langle y^2+1 \rangle$ in $\mathbb{R}[x,y]$. The quotient ring is isomorphic to $\mathbb{C}[x]$, which has Krull dimension one, but there are no points $(a_1,a_2) \in \mathbb{R}^2$ with $a_2^2+1 = 0$.
