So, the problem is:
Show that the map $\iota: k[x_1, \dots, x_n] \to \overline{k}[x_1, \dots, x_n]$ induces a map of sets $\pi: \text{mSpec} \left( \overline{k}[x_1, \dots, x_n] \right) \to \text{mSpec} \left( k[x_1, \dots, x_n] \right)$
The only natural map I can think of is taking preimages:
$M \to \iota^{-1} \left( M \right)$. The main issue I'm having is proving that $\iota^{-1} \left( M \right)$ is indeed maximal. My hope was that I could show that $k[x_1, \dots, x_n] \to \overline{k}[x_1, \dots, x_n]/M $ is surjective. I thought this might be true in general since it holds for the (extremely limited) case $\mathbb{R}[x] \to C[x]/(x-i)$.
My plan for trying to prove this in general was to use the fact that $M = (x_1 - a_1, \dots, x_n - a_n)$, and somehow use the fact that $a_1, \dots, a_n$ are all algebraic over $k$
This route is not leading to any natural insights, so I'm pretty stumped. Hints? Solutions?
I just realized the comment I had basically answers the question: for $M=(x_1-a_1,\dots,x_n-a_n)$ consider the composition
$$k[x_1,\dots,x_n]\overset{\iota}{\to} \overline k[x_1,\dots,x_n]\to k[x_1,\dots,x_n]/M\overset{\simeq}{\to}\overline k,$$
explicitly given by $f\mapsto f(a_1,\dots,a_n)$. The image of this composition is exactly $k[a_1,\dots,a_n]$, which is a field (if you are not aware of why then I can add a proof), so the kernel (which is exactly $\iota^{-1}(M)$) is a maximal ideal of $k[x_1,\dots,x_n]$.