Let $I$ be the ideal of $R$ generated by all the monomials of degree at least $2$, that is, $$I=\left\langle x_{i_1}^{e_1}\cdots x_{i_r}^{e_r},:, r \geq 1, e_j \geq 2\right\rangle.$$ We claim that $I$ is a maximal ideal which is not of the form $\mathrm{m}{\mathrm{a}}$ for any $\mathbf{a} \in \prod{i=1}^{\infty} k$.

First, we show that $I$ is a proper ideal. Suppose for contradiction that $I=R$. Then every polynomial $f \in R$ can be written as a linear combination of monomials of degree at most $1$. But this is impossible since we have infinitely many variables $x_1, x_2, \ldots$, so there are infinitely many monomials of degree at least $2$. Hence $I$ is proper.

Next, we show that $I$ is a maximal ideal. Let $J$ be an ideal of $R$ strictly containing $I$. We will show that $J=R$, which implies that $I$ is maximal.

Let $f \in J \setminus I$. Then $f$ must be a monomial of degree at most $1$. Let $x_i$ be a variable that appears in $f$. Since $f \notin I$, we have $x_i \notin \operatorname{supp}(f)$, where $\operatorname{supp}(f)$ denotes the support of $f$, that is, the set of variables appearing in $f$ with nonzero coefficient. Hence we can write $$f=x_i g + h,$$ where $g,h \in R$ and $h$ does not involve $x_i$. Since $f \in J$ and $J$ contains $I$, we have $x_i g \in J$ and $h \in J$. But $J$ strictly contains $I$, so there exists some monomial $m \in J$ which is not in $I$. Since $m$ is not a monomial of degree at least $2$, it must be a monomial of degree $1$, say $m=x_j$ for some $j$. But then $x_j \in J$, so $x_i g + x_j \in J$. Repeating this argument for all variables appearing in $f$, we conclude that $f \in J$. Hence $J=R$, as desired.

Finally, we show that $I$ is not of the form $\mathrm{m}{\mathrm{a}}$ for any $\mathbf{a} \in \prod{i=1}^{\infty} k$. Suppose for contradiction that $I=\mathrm{m}{\mathrm{a}}$ for some $\mathbf{a} \in \prod{i=1}^{\infty} k$. Then $\mathrm{a}$ must be the zero sequence, since $I$ contains all monomials of degree at least $2$. But then $\mathrm{m}{\mathrm{a}}$ is not maximal, since it is properly contained in $\left\langle x_1\right\rangle$, which is maximal. This is a contradiction, so $I$ is not of the form $\mathrm{m}{\mathrm{a}}$ for any $\mathbf{a} \in \prod_{i=1}^{\infty} k$.

Let $Q$ be a quadric in $\mathbb{P}^{n-1}$ and $H$ a hyperplane in $\mathbb{P}^{n-1}$ such that $Q\cap H$ is a smooth quadric of dimension $n-2$. We want to define a map $\phi:(Q\cap H)\setminus {x}\rightarrow Q_{n-2}$ such that each fiber of $\phi$ is isomorphic to $\mathbb{A}^1$.

Let $x$ be a point in $Q\cap H$. Then we can identify $T_x(Q\cap H)$ with the orthogonal complement of $TxH$ in $Tx\mathbb{P}^{n-1}$. Since $Q$ is smooth and contains $x$, we have $T_xQ=T_xH\oplus T_x(Q\cap H)$.

Now consider the projection map $\pi:T_xQ\rightarrow T_x(Q\cap H)$. Since $Q$ is a quadric, $\pi$ is an isomorphism. Let $S_x$ denote the set of lines in $T_xQ$ passing through $x$. We can identify $S_x$ with the set of points in $Q$ that are incident with $x$.

Let $v\in T_x(Q\cap H)$ be a nonzero vector. We define a map $\psi_v:S_x\rightarrow\mathbb{A}^1$ as follows. For each $p\in S_x$, let $w_p$ be the unique vector in $T_pQ$ such that $w_p\cdot v=0$ and $|w_p|=1$. We can define $\psi_v(p)$ to be the unique $t\in\mathbb{A}^1$ such that $x+tv+w_p\in Q$.

Note that $\psi_v$ is well-defined because $Q$ is a quadric, so $w_p$ is uniquely determined. Moreover, $\psi_v$ is a bijection, since $Q$ is smooth and contains no lines, so the map $Q\rightarrow \mathbb{P}(T_xQ)$ given by $p\mapsto [x,p]$ is a local isomorphism at $x$.

Finally, we define $\phi:(Q\cap H)\setminus {x}\rightarrow Q_{n-2}$ by $\phi(p)=\psi_v(p)$, where $v$ is any nonzero vector in $T_x(Q\cap H)$ that is orthogonal to $TpH$. This map is well-defined and has the desired property that each fiber is isomorphic to $\mathbb{A}^1$.Regenerate response

Assume that $\operatorname{dim}(Z) = \operatorname{dim}(X) = n$. We want to show that $Z$ is a component of $X$.

Recall that a component of $X$ is a maximal irreducible closed subset of $X$. Thus, we need to show that $Z$ is irreducible and maximal among irreducible closed subsets of $X$.

To show that $Z$ is irreducible, we need to show that if $Z = Z_1 \cup Z_2$ for some closed subsets $Z_1$ and $Z_2$ of $X$, then $Z_1 = Z$ or $Z_2 = Z$. Since $Z$ is irreducible, it is enough to show that $Z_1$ or $Z_2$ has dimension $n$. Without loss of generality, assume that $\operatorname{dim}(Z_1) \geq n$. Then there exists a chain of irreducible closed subsets of $Z_1$ with lengths $n$, say $Z_1^0 \subsetneq Z_1^1 \subsetneq \cdots \subsetneq Z_1^n$. By the Noether normalization lemma, there exists a finite surjective morphism $\phi:X\rightarrow \mathbb{A}^n$ such that $\phi(Z_1^n)$ is closed in $\mathbb{A}^n$. Let $Y = \phi^{-1}(\phi(Z_1^n))$. Since $\phi$ is surjective, $Y$ is closed in $X$, and since $\phi$ is a finite morphism, $Y$ is irreducible. Moreover, $\operatorname{dim}(Y) = \operatorname{dim}(Z_1^n) = n$, so $Y$ contains $Z$. But $Y \subseteq Z_1$, so $Z_1 = X$. Therefore, $Z$ is irreducible.

To show that $Z$ is maximal among irreducible closed subsets of $X$, let $Z’$ be an irreducible closed subset of $X$ such that $Z \subseteq Z’$. We need to show that $Z = Z’$. Since $\operatorname{dim}(Z’) = n$, there exists a chain of irreducible closed subsets of $Z’$ with lengths $n$, say $Z’^0 \subsetneq Z’^1 \subsetneq \cdots \subsetneq Z’^n$. By the Noether normalization lemma, there exists a finite surjective morphism $\phi:X\rightarrow \mathbb{A}^n$ such that $\phi(Z’^n)$ is closed in $\mathbb{A}^n$. Let $Y = \phi^{-1}(\phi(Z’^n))$. Since $\phi$ is surjective, $Y$ is closed in $X$, and since $\phi$ is a finite morphism, $Y$ is irreducible. Moreover, $\operatorname{dim}(Y) = \operatorname{dim}(Z’^n) = n$, so $Y$ contains $Z’$. But $Z’ \subseteq Z$, so $Y$ contains $Z$. Therefore, $Z = Y \subseteq Z’$, and hence $Z = Z’$.

Therefore, $Z$ is a component of $X$.