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In this answer I will treat the case in which $|\text{ }|$ is not discrete.

I first claim that $\mathfrak m_0$ is not the restriction of any proper ideal in $k^{\infty}.$ Indeed, choose $x \in k$ such that $0 < |x| < 1$. Then $(x^i)$ is an element of $\mathfrak m_0$ which is invertible in $k^{\infty}$ (with inverse equal to $(x^{-i})$, and so $\mathfrak m_0$ generates the unit ideal of $k^{\infty}$.

This doesn't contradict anything; the maximal ideals of $k^{\infty}$ pull-back to prime ideals in $\mathcal C(k)$ which are simply no not maximal (as often happens with maps of rings).

Furthermore, this pull-back is injective.

To see this, we first introduce some notation; namely, we let $\mathfrak m_{\mathcal U}$ denote the prime ideal of $k^{\infty}$ corresponding to the non-principal ultra-filter ${\mathcal U}$,and recall that $\mathfrak m_{\mathcal U}$ is defined as follows: an element $(x_i)$ lies in $\mathfrak m_{\mathcal U}$ if and only if ${i \{i \, | \, x_i = 0}$ 0\}$lies in in$\mathcal U$. Now suppose that$\mathcal U_1$and$\mathcal U_2$are two distinct non-principal ultra-filters. Let$A$be a set lying in$\mathcal U_1$, but not in$\mathcal U_2$. Then$A^c$, the complement of$A$, lies in$\mathcal U_2$. Choose$x \in k$such$0 < | x | < 1,$and let$x_i = x^i$if$i \in A$and$x_i = 0$if$i \not\in A$. Then$(x_i)$is an element of$\mathcal C(k)$, in fact of$\mathfrak m_0$, and it lies in$\mathfrak m_{\mathcal U_2}$but not in$\mathfrak m_{\mathcal U_1}$. Thus$\mathfrak m_{\mathcal U_1}$and$\mathfrak m_{\mathcal U_2}$have distinct pull-backs. So the map${\mathrm Spec } k^{\infty} $k^{\infty} \rightarrow {\mathrm$ Spec } \mathcal $\mathcal C(k)$ is injective and dominant (since it comes from an injective map of rings), but is not surjective. Choosing the valuation $|\text{ }|$ allows us to add to Spec $k^{\infty}$ (which is the Stone-Cech compactification of $\mathbb Z_+$) an extra point dominating all the other points at infinity (i.e. all the non-principal ultrafilters), because the valuation now gives us a definitive way to compute limits (provided we begin with a Cauchy sequence).

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In this answer I will treat the case in which $|\text{ }|$ is not discrete.

I first claim that $\mathfrak m_0$ is not the restriction of any proper ideal in $k^{\infty}.$ Indeed, choose $x \in k$ such that $0 < |x| < 1$. Then $(x^i)$ is an element of $\mathfrak m_0$ which is invertible in $k^{\infty}$ (with inverse equal to $(x^{-i})$, and so $\mathfrak m_0$ generates the unit ideal of $k^{\infty}$.

This doesn't contradict anything; the maximal ideals of $k^{\infty}$ pull-back to prime ideals in $\mathcal C(k)$ which are simply no maximal (as often happens with maps of rings).

Furthermore, this pull-back is injective.

To see this, we first introduce some notation; namely, we let $\mathfrak m_{\mathcal U}$ denote the prime ideal of $k^{\infty}$ corresponding to the non-principal ultra-filter ${\mathcal U}$,and recall that $\mathfrak m_{\mathcal U}$ is defined as follows: an element $(x_i)$ lies in $\mathfrak m_{\mathcal U}$ if and only if ${i \, | \, x_i = 0}$ lies in in $\mathcal U$.

Now suppose that $\mathcal U_1$ and $\mathcal U_2$ are two distinct non-principal ultra-filters. Let $A$ be a set lying in $\mathcal U_1$, but not in $\mathcal U_2$. Then $A^c$, the complement of $A$, lies in $\mathcal U_2$.
Choose $x \in k$ such $0 < | x | < 1,$ and let $x_i = x^i$ if $i \in A$ and $x_i = 0$ if $i \not\in A$. Then $(x_i)$ is an element of $\mathcal C(k)$, in fact of $\mathfrak m_0$, and it lies in $\mathfrak m_{\mathcal U_2}$ but not in $\mathfrak m_{\mathcal U_1}$.

Thus $\mathfrak m_{\mathcal U_1}$ and $\mathfrak m_{\mathcal U_2}$ have distinct pull-backs.

So the map ${\mathrm Spec} k^{\infty} \rightarrow {\mathrm Spec} \mathcal C(k)$ is injective and dominant (since it comes from an injective map of rings), but is not surjective. Choosing the valuation $|\text{ }|$ allows us to add to Spec $k^{\infty}$ (which is the Stone-Cech compactification of $\mathbb Z_+$) an extra point dominating all the other points at infinity (i.e. all the non-principal ultrafilters), because the valuation now gives us a definitive way to compute limits (provided we begin with a Cauchy sequence).