As a complement to the excellent answers already given by Mariano and Qiaochu, let me give a description and two illustrations of the closed points of $\mathbb A^n_k$.
First take an algebraic closure $\bar k$ of $k$.
The closed points of $\mathbb A^n_k$ are the maximal ideals of the polynomial ring $k[x_1, ..., x_n]$.
Now, given an $n$-tuple $a=(a_1, ..., a_n)\in \bar k^n$, we can associate to it the set $\mathfrak m_a$ of polynomials $P\in k[x_1, ..., x_n]$ vanishing at $a$ (i.e. $P(a_1, ..., a_n)=0$).
Then $\mathfrak m_a$ is a maximal ideal of $k[x_1, ..., x_n]$ and the pleasant surprise is that we obtain all the maximal ideals by this procedure.
Moreover, we know for which $n$-tuples $a,b\in \bar k^n$ we'll have $\mathfrak m_a=\mathfrak m_b$: that will be the case if and only if there exists an automorphism $\sigma\in Aut_k(\bar k)$ of $\bar k$ fixing $k$ such that $\sigma (a)=b$ (meaning $\sigma (a_i)=b_i$ for $i=1, ...,n$)
Let me illustrate with the smallest example: the affine line $\mathbb A^1_{\mathbb F_2}$ over the field $\mathbb F_2$ with $2$ elements.
Its closed points are given by the elements $ a\in \overline {\mathbb F_2}$ of an algebraic closure of $\mathbb F_2$ and two such elements give the same maximal ideal iff they have the same minimal polynomial. (Equivalently the closed points are given by the irreducible polynomials in $\mathbb F_2[X]$).
So despite apperances $\mathbb A^1_{\mathbb F_2}$ has infinitely many closed points!
(And, for the record, one non closed point: its so-called generic point)
As another illustration I'll describe the closed points of $\mathbb A^2_{\mathbb R}=Spec(\mathbb R[x,y])$ with the help of the group $Aut_{\mathbb R}( \mathbb C)=\lbrace I, \sigma\rbrace$ generated by complex conjugation $\sigma:z\mapsto \bar z$.
$\bullet$ First we have the classical maximal ideals $\langle x-r,y-s \rangle (r,s\in \mathbb R)$ corresponding to $(r,s)\in \mathbb R^2$ $ \bullet \bullet $ The other closed points correspond to the less classical maximal ideals $\mathfrak m_{(z,w)}$ associated to $(z,w)\in \mathbb C^2 \setminus \mathbb R^2$.
If $z=a+bi, w=c+di$ a little calculation shows that $m_{(z,w)}=m_{(\bar z,\bar w)}=\langle dx-by-ad+bc, (x-a)^2+b^2, (y-c)^2+d^2\rangle \subset \mathbb R[x,y]$ (One of the last two generators is redundant: if, say, $d\neq0$ keep the last one and dismiss the second. If both $b,d\neq0$, arbitrarily dismiss one of the last two generators)
For example $m_{(1,2-3i)}=\langle -3x+3,(x-1)^2+0^2,(y-2)^2+3^2 \rangle =\langle -x+1,y^2-4y+13 \rangle $
Edit
Although there are a gazillion books describing the Hilbert Nullstellensatz, the description I give at the beginning of the maximal ideals of $k[x_1, ..., x_n]$ is not so easy to find in the literature.
As always Bourbaki comes to the rescue: Commutative Algebra, Chapter V, §3.4, Proposition 2, page 351.