Remember that in a commutative ring with unity $R$, you have the following condition for maximal ideals:
An ideal $I$ of $R$ is maximal $\iff R/I$ is a field
Then if $R = k[x_1, \dots, x_n]$, you're considering the map $ \phi: k[x_1, \dots, x_n] \to k $ given by $\phi(f(x_1, \dots, x_n)) = f(0, \dots, 0)$, that is, the evaluation at zero map. Since this map is clearly surjective, by the first isomorphism theorem you get
$ k[x_1, \dots, x_n]/\ker{(\phi)} \cong k $
But then since $\ker{(\phi)} = \langle x_1, \dots, x_n \rangle$ is the ideal of all polynomials in $k[x_1, \dots, x_n]$ with zero constant term, and since $k$ is a field, by the above stated result, this shows that the ideal is maximal.