Yes, you can say this.
Let us suppose $K[x] \cap P$ is nontrivial. Since $K$ is algebraically closed and $P$ is prime, there is some $a \in K$ such that $x - a \in P$.
And if $K[y] \cap P$ is also nontrivial, then there is some $b \in K$ such that $y - b \in P$.
Let $R = K[x, y] / (x - a, y - b) \cong K$, and let $\pi : K[x, y] \to R$ be the canonical map. Then $R / \pi(P) \cong K[x, y]/P$.
Now since $P$ is prime, we see that $K[x, y] / P \cong R/\pi(P)$ is an integral domain. Therefore, $\pi(P)$ is prime, hence maximal. Then $P$ is maximal.
Obviously, this generalises to polynomials over any finite number of variables.