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Let $k$ be an infinite field, not algebraically closed. Let $\operatorname{Specm} k[x_1,\cdots,x_n]$ be the set of maximal ideals of $k[x_1,\cdots,x_n]$ and define a topology on $\operatorname{Specm} k[x_1,\cdots,x_n]$ in which the closed sets are of the form $Z(E) = \left\{m \in \operatorname{Specm} k[x_1,\cdots,x_n]: E \subseteq m \right\}$, where $E$ is a subset of $k[x_1,\cdots,x_n]$. Now, let $X$ be the subset of $\operatorname{Specm} k[x_1,\cdots,x_n]$, consisting of maximal ideals of the form $(x_1-c_1,\cdots,x_n-c_n)$ with $c_i \in k$. I need a hint towards showing that $X$ is dense in $\operatorname{Specm} k[x_1,\cdots,x_n]$. I proved this for the case where $k=\mathbb{R}$ and $n=1$, but i made decisive use of the fact that $\mathbb{R}[x]$ is a principal ideal domain. The case $\mathbb{R}[x_1,x_2]$ seems to be fundamentally different.

PS: Please, only a hint, not the solution. Thanks.

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Here is a proof i came up with: Let $m \in \operatorname{Specm}k[x_1,\cdots,x_n]$ and let $U_m$ be a neighborhood of $m$. Then by definition $U_m=\left\{m' \in \operatorname{Specm}k[x_1,\cdots,x_n] : m' \nsupseteq \alpha \right\}$ for some ideal $\alpha$. Since $k[x_1,\cdots,x_n]$ is a Noetherian ring, $\alpha=(f_1,\cdots,f_k) , \, f_i \in k[x_1,\cdots,x_n]$. Proceed by contradiction, i.e. suppose that for any $r_1,r_2 \in k$ we have that $(x-r_1,x-r_2) \supseteq \alpha$. Then $f_i(r_1,r_2)=0, \, \forall r_1,r_2 \in k$. Then $f_i$ is zero on the zero set of the zero ideal $(0)$ and by Hilbert's Nullestellensatz $f_i^r \in (0)$ for some positive integer $r$. Hence $f_i^r=0$ and since $k[x_1,\cdots,x_n]$ is an integral domain, $f_i=0$. Thus $\alpha=0$, which implies that $U_m=\emptyset$. This is a contradiction, since we must have that $m \in U_m$. Hence there exist $r_1,r_2 \in k$ such that $(x-r_1,x-r_2) \in U_m$.

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