= Solution
The variable $t$ is central. To prove the <leading-coefficient proof for a central polynomial extension>, let $I$ be any <right ideal> of $R[t]$ and set
$$
L_d=\left\{a_d:\sum_{i=0}^d a_it^i\in I\right\}\subseteq R.
$$
Zero leading coefficients are allowed. Addition and right multiplication by constants show that $L_d$ is a <right ideal>. Multiplication by $t$ gives $L_d\subseteq L_{d+1}$. Since $R$ is a <right Noetherian ring>, this chain stabilizes, say at $L_N$, and each $L_d$ for $d\le N$ has finitely many generators $a_{dj}$. Choose $f_{dj}\in I$ of degree at most $d$ lifting these coefficients.
These finitely many polynomial lifts generate $I$ on the right. Indeed, for $f\in I$ of degree $m$, let $d=\min(m,N)$. Write its leading coefficient as $\sum_j a_{dj}r_j$. Then
$$
f-\sum_j f_{dj}\,r_jt^{m-d}
$$
belongs to $I$ and has degree less than $m$, because $L_m=L_d$ when $m>N$. Induction on degree reduces $f$ to a right linear combination of the chosen lifts. This proves finite generation of every <right ideal>, and hence
$$
\boxed{R\text{ right Noetherian}\Longrightarrow R[t]\text{ right Noetherian}.}
$$
Centrality matters: it is what permits the displayed leading-coefficient cancellation without twisting the coefficients.
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