[EDIT (Feb. 27, 2024): No answer to the reference question yet, but I explain a more general statement at the end.]
Let $k$ be a field and let $\underline{a}=(a_0,\dots,a_n)$ be a tuple of positive integers. Denote by $P_\underline{a}$ the corresponding weighted projective space over $k$, i.e. the quotient of $U:=\mathbb{A}^{n+1}_k\smallsetminus\{0\}$ by the $\mathbb{G}_m$-action with weight $\underline{a}$.
Of course, every nonzero $(x_0,\dots,x_n)\in k^n$ determines a $k$-rational point in $P_\underline{a}$.
Fact: These are all the $k$-rational points, i.e. the quotient map $\pi:U\to P_\underline{a}$ induces a surjection on $k$-rational points.
This fact is not immediately obvious from the definition. Clearly it extends to $R$-valued points if $R/k$ is a field extension, but not if $R$ is an arbitrary $k$-algebra, not even local or artinian (so it's not ``just Hilbert 90'').
The proof is simple once you see it (see below), but my question is:
Question: Does anyone have a reference?
[EDIT: same question for the proposition below.]
Now here is a (sketch of) proof: We may assume that $a_0,\dots,a_n$ are relatively prime. Fix integers $u_i$ such that $\sum_i u_ia_i=1$. Let $W\subset P_\underline{a}$ be the open subscheme where all coordinates are nonzero, and $V\subset U$ be defined by $\prod_i x_i^{u_i}=1$. Then one checks that $\pi$ induces an isomorphism from $V$ to $W$, so all points of $W(k)$ lift to $U$. Now, $P_\underline{a}\smallsetminus W$ is a finite union of $(n-1)$-dimensional weighted projective spaces, so we conclude by induction on $n$.
[EDIT Feb. 27, 2024]: So here is a more general fact, with an elementary proof without mention of GIT or Hilbert 90.
Let $S=\bigoplus_{n≥0}S_n$ be a positively graded ring (no finiteness conditions!). Put $X=\mathrm{Proj}(S)$, $C=\mathrm{Spec}(S)\smallsetminus\mathrm{Spec}(S_0)$ (the punctured cone), and $\pi:C\to X$ the natural projection.
Proposition. For any field $k$, the map $C(k)\to X(k)$ induced by $\pi$ is surjective.
More precisely, for any $x:\mathrm{Spec}(k)\to X$, the reduced fiber $(C\times_{\pi,X,x}\mathrm{Spec}(k))_\mathrm{red}$ is isomorphic to $\mathrm{Spec}(k[t,t^{-1}])$.
Proof. Any $x$ as in the statement factors through some $D_+(f)\subset X$ for some $f\in S_d$, $d>0$. Now $\pi^{-1}(D_+(f))=\mathrm{Spec}(S[1/f])$ (a $\mathbb{Z}$-graded ring) and $D_+(f)=\mathrm{Spec}(S_{(f)})$ where $S_{(f)}$ is the degree zero part of $(S[1/f])$. Applying the base change $x$ (and checking compatibilities) the fiber in question is Spec of the $\mathbb{Z}$-graded $k$-algebra $S[1/f]\otimes_{S_{(f)}}k$. So everything follows from:
Lemma. Let $B:=\bigoplus_{n\in\mathbb{Z}}B_n$ be a $\mathbb{Z}$-graded ring, and let $d>0$. Assume $B_d$ has an element $f$ invertible in $B$. Then:
(1) The morphism $\phi_f: B_0[t,t^{-1}]\to B^{(d)}=\bigoplus_{n\in d\mathbb{Z}}B_n$ sending $t$ to $f$ is an isomorphism.
(2) Assume $B_0$ is a field and $d$ is minimal (such that $B_d\cap B^\times\neq\emptyset$). Then every $g\in B_e$ with $e\notin d\mathbb{Z}$ satisfies $g^d=0$.
In particular, the composite $B_0[t,t^{-1}]\to B^{(d)}\hookrightarrow B \to B_\mathrm{red}$ is an isomorphism.
Proof. (1) is immediate; more generally, multiplication by $f^m$ ($m\in\mathbb{Z}$) induces isomorphisms $B_e\cong B_{e+md}$. By the minimality assumption and Euclidean division it follows that in (2), $g$ is not invertible. Now $g^d\,f^{-e}\in B_0$ which is a field, whence $g^d=0$.
