Two questions on Hilbert's Nullstellensatz. We have this strong version of Hilbert's Nullstellensatz:
If $k$ is any field, every maximal ideal of $k[x_1, \dots, x_n]$ has residue field a finite extension of $k$. 
Now I have two questions: 

1) Why this is equivalent to say that any field extension of $k$ that is finitely generated as ring is finitely generated as a module? 
2) And why this form implies the weak form of Nullstellensatz? (If $k$ is an algebraically closed field, then the maximal ideals of $k[x_1, \dots, x_n]$ are precisely those ideals of the form $(x_1-\alpha_1, \dots, x_n -\alpha_n)$).

 A: A lot of this just echoes what Alex Youcis has already said; I thought it would be good to have an answer.
What does it mean for a $B$-algebra $B \to A$ to be finitely generated (or "finite type")? The translation between this statement and "$A$ is a quotient of a polynomial ring $B[x_1, \dots, x_n]$ in finitely many variables by an ideal $I$" must become second nature. Of course, explicitly writing $A$ out in this way gives something slightly more: a choice of generators and a realization of $\operatorname{Spec} A$ inside $\mathbf{A}^n_B$ as a "closed subscheme". Now, $A$ is a field if and only if $I$ is maximal.
If $k$ is algebraically closed then you should be able to show that it has no nontrivial finite field extensions. Maybe it's good to turn things around and see where ideals like $\mathfrak{m} = (x_1 - a_1, \dots, x_n - a_n)$ turn up. What follows applies to any $k$ and describes the "$k$-rational points" of $\mathbf{A}^n_k$. The difference will be that there are closed points not of this form, e.g., the maximal ideal $(x^2 + 1, y - x)$ of $\mathbf{Q}[x, y]$.
Say I have a point $(a_1, \dots, a_n)$ in $k^n$. This induces a morphism of $k$-algebras $\xi\colon k[x_1, \dots, x_n] \to k$, $f \mapsto f(a_1, \dots, a_n)$ and the kernel is $\mathfrak{m}$. Work this out! Now show yourself that you can recover the point of $k^n$ from $\xi$.
