There is a very ampleness condition due to Delorme in [Del75a]. You can also look at [BR86] for an English reference. We have changed the dimension $n$ in your notation to an $r$ to closer match the notation in these references.
We start with some definitions.
Definition [Del75a, Def. 2.1; BR86, Defs. 4B.1, 4B.2, and 4B.3]. We let $m_J := \operatorname{lcm}\{a_i \mid i \in J\}$ for every non-empty subset $J \subseteq \{0,1,\ldots,r\}$, and let $m := m_{\{1,2,\ldots,r\}}$. Now set
$$G := \begin{cases}
-a_r & \text{if $r = 0$, and}\\
\displaystyle-\sum_{i=0}^r a_i + \frac{1}{r}\sum_{2 \le \nu \le r+1} \binom{r-1}{\nu-2}^{-1} \sum_{\lvert J \rvert = \nu} m_J & \text{otherwise.}
\end{cases}$$
We then say that an integer $n \ge 0$ satisfies condition $D(n)$ if one of the following equivalent conditions hold:
Given a relation $\sum_{i = 0}^r B_ia_i = n + km$ with $k \in \mathbf{Z}_{>0}$ and $B_i \in \mathbf{Z}_{\ge0}$ for every $i$, there exist $b_i \in \mathbf{Z}_{\ge0}$ with $B_i \ge b_i$ for every $i$, such that $\sum_{i = 0}^r b_ia_i=km$.
Every monomial $\prod_{i=0}^r x_i^{B_i}$ of degree $n+km$ is divisible by a monomial $\prod_{i=0}^r x_i^{b_i}$ of degree $km$.
We then define $F$ to be the smallest integer such that $n > F$ implies $D(n)$ holds. We also define $E$ to be the smallest integer such that $n > E$ implies $D(mn)$ holds. Note that $mE \le F$.
One can then show that $F$ is finite and $F \le G$ [Del75a, Prop. 2.2; BR86, Prop. 4B.5]. The proof is a double induction on $k$ and $r$, and boils down to the pigeon-hole principle.
With notation as above, we then have the following:
Theorem [Del75a, Prop. 2.3; BR86, Thm. 4B.7]. Let $X = \mathbf{P}(a_0,a_1,\ldots,a_r)$ be a weighted projective space over a commutative ring $A$, where $a_i \ge 1$ for every $1$. With notation as above, we have the following:
$\mathcal{O}_X(m)$ is an ample invertible sheaf.
If $n > F$, then the sheaf $\mathcal{O}_X(n)$ is globally generated.
If $n > 0$ and $n > E$, then the sheaf $\mathcal{O}_X(nm)$ is very ample.
We prove (3), since this is what you are interested in. For every $p \in \mathbf{Z}_{>0}$, the condition $D(mn)$ implies that every monomial of degree $pmn = (p-1)mn + mn$ in $A[x_0,x_1,\ldots,x_r]$ is divisible by a monomial of degree $(p-1)mn$. Thus, the $mn$th Veronese subring $A[x_0,x_1,\ldots,x_r]^{(mn)}$ of $A[x_0,x_1,\ldots,x_r]$ is generated in degree $1$ over $A$.
You can therefore embed $X$ into $\mathbf{P}^N_A$, where $N$ is the number of generators of $A[x_0,x_1,\ldots,x_r]_{mn}$, by
$$X = \operatorname{Proj}\bigl(A[x_0,x_1,\ldots,x_r]\bigr) \simeq \operatorname{Proj}\bigl(A[x_0,x_1,\ldots,x_r]^{(mn)}\bigr) \hookrightarrow \mathbf{P}^N_A.$$
Unraveling the definitions given above, we note that in particular, setting
$$n = \biggl\lfloor\frac{1}{m}G\biggr\rfloor+1$$
works.
References
[BR86] Mauro Beltrametti and Lorenzo Robbiano, "Introduction to the theory of weighted projective spaces," Exposition. Math. 4 (1986), no. 2, 111–162. mr: 879909.
[Del75a] Charles Delorme, "Espaces projectifs anisotropes," Bull. Soc. Math. France 103 (1975), no. 2, 203–223. doi: 10.24033/bsmf.1802. mr: 404277.
[Del75b] Charles Delorme, "Erratum: 'Espaces projectifs anisotropes'," Bull. Soc. Math. France 103 (1975), no. 4, 510. doi: 10.24033/bsmf.1812. mr: 404278.
Best Answer
$\operatorname{Proj}(A[x_0,...x_n])$ is not a "projective space" in the sense of synthetic geometry. If $k$ is an $A$-algebra that is a field, then the set of $k$-points of $\operatorname{Proj}(A[x_0,...x_n])$ (that is, the morphisms $\operatorname{Spec} k\to \operatorname{Proj}(A[x_0,...x_n])$ over $\operatorname{Spec} A$) is a projective space in the sense of synthetic geometry, since it can be naturally identified with the projective space of the vector space $k^{n+1}$. Explicitly, a point $(a_0,\dots,a_n)\in k^{n+1}$ determines a homomorphism of graded $A$-algebras $A[x_0,\dots,x_n]\to k[t]$ which maps each $x_i$ to $a_it$, and this determines a morphism $\operatorname{Spec} k\cong \operatorname{Proj}(k[t])\to\operatorname{Proj}(A[x_0,...x_n])$. It can be shown that this morphism is unchanged if you multiply $(a_0,\dots,a_n)$ by a scalar, and that every $k$-point of $\operatorname{Proj}(A[x_0,...x_n])$ arises in this way, so that $k$-points of $\operatorname{Proj}(A[x_0,...x_n])$ are naturally in bijection with lines through the origin in $k^{n+1}$.
To be explicit, the set of lines through the origin in $k^{n+1}$ forms a projective space in the sets of synthetic geometry in the following way. A "point" is a line through the origin in $k^{n+1}$, and a "line" is a plane through the origin in $k^{n+1}$. The incidence relation is just containment: a "point" is on a "line" if the corresponding line through the origin is a subset of the corresponding plane through the origin.