Showing there are at least one and only finitely many maximal ideals containing the extension of a maximal ideal

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Let $F$ be a field and $M$ a maximal ideal of $F[x_1, x_2, ..., x_n]$. Let $K$ be an algebraic closure of $F$. Show that $M$ is contained in at least one and in only finitely many maximal ideals of $K[x_1, x_2, ..., x_n]$.

Any hints to get me started would be much appreciated.

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Think about residue fields. By the Nullstellensatz, the residue field $L=F[x_1,\dots,x_n]/M$ is a finite extension of $F$, and the residue field of $K[x_1,\dots,x_n]$ at any maximal ideal is $K$. From this, you can show that there is a bijection between maximal ideals extending $M$ and embeddings of $L$ into $K$ (extending the given embedding $F\to K$).