This problem is from a qualifying exam.
I am to show that if $k$ is a field and if $n > 1$, then no maximal ideal of $k[x_1, \dots, x_n]$ is principal.
I understand why this is true if $k$ is algebraically closed, since every maximal ideal is of the form $(x_1 - a_1, \dots, x_n - a_n)$ for some $a_i \in k$.
My attempt for $k$ an arbitrary field. Let $m \subset k[x_1, \dots, x_n]$ be a maximal ideal and let $K$ be the algebraic closure of $k$. Let $M \subset K[x_1, \dots, x_n]$ be the push-forward ideal of $m$, namely, $M = K[x_1, \dots, x_n]m$. Assume $M$ is not the unit ideal, and so is contained in the maximal ideal $(x_1-a_1, \dots, x_n - a_n)$ of $K[x_1, \dots, x_n]$ for some $a_i \in K$.
Suppose $m$ is principal and generated by some irreducible $f \in k[x_1, \dots, x_n]$. Observe that the subset $A$ of $(x_1 - a_1, \dots, x_n - a_n)$ consisting of all polynomials whose coefficients are in $k$ is an ideal of $k[x_1, \dots, x_n]$ containing $m$, and therefore $A = m$ by maximality. For each $i$, let $f_i \in k[x_1, \dots, x_n]$ be a polynomial of minimal degree such that $x_i - a_i$ divides $f_i$ in $K[x_1, \dots, x_n]$ -- such a polynomial exists since each $a_i$ is algebraic over $k$. In particular, $f_i \in A = m$. Since $f$ divides $f_i$ in $k[x_1, \dots, x_n]$, it follows by the minimality of $f_i$ that $f_i = f$ for all $i$. We gather $x_i - a_i$ divides $f$ for all $i$.
Note that if any of the $a_i$ are in $k$ then we are done because $f = f_i = x_i - a_i$, again by the minimality of $f_i$, which is a contradiction since $x_j - a_j \nmid x_i - a_i$ for $i \neq j$.
However, I cannot find the contradiction in the case all the $a_i$ are in $K - k$. Does anyone see how to proceed along this particular route? If not, I welcome other suggestions. I understand there are arguments using the Krull dimension, but given that Krull dimension is not covered in our algebra course, I'm hesitant to use such an argument on an exam. I think it is intended we use Nullstellensatz, but I'm not entirely sure.
EDIT: As discussed in the comments, one also must account for the case $M$ is unital, if this strategy is to be kept.