This answer is a polished version of what I wrote in the comments.
Let me first restate Weyl's theorem in a modern language. We fix an
infinite field $\mathbb{K}$ (for example, $\mathbb{Q}$). Its elements will
be called scalars.
Theorem 1 (Weyl's principle of irrelevance of algebraic inequalities). Let
$n$ and $m$ be nonnegative integers. Let $\mathcal{P}$ be the polynomial ring
$\mathbb{K}\left[ x_{1},x_{2},\ldots,x_{n}\right] $. Let $F,R_{1}
,R_{2},\ldots,R_{m}\in\mathcal{P}$ be polynomials such that $R_{1}
,R_{2},\ldots,R_{m}$ are nonzero. Assume that every $n$-tuple $\left(
a_{1},a_{2},\ldots,a_{n}\right) \in\mathbb{K}^{n}$ of scalars that satisfies
\begin{equation}
R_{i}\left( a_{1},a_{2},\ldots,a_{n}\right) \neq0\qquad\text{for all }
i\in\left\{ 1,2,\ldots,m\right\}
\label{darij1.eq.t1.1}
\tag{1}
\end{equation}
also satisfies
\begin{equation}
F\left( a_{1},a_{2},\ldots,a_{n}\right) =0.
\label{darij1.eq.t1.2}
\tag{2}
\end{equation}
Then, $F=0$.
We shall derive this from the following two known facts:
Theorem 2. Let $n$ be a nonnegative integer. Let $\mathcal{P}$ be the
polynomial ring $\mathbb{K}\left[ x_{1},x_{2},\ldots,x_{n}\right] $. Then,
$\mathcal{P}$ is an integral domain.
Theorem 3. Let $n$ be a nonnegative integer. Let $\mathcal{P}$ be the
polynomial ring $\mathbb{K}\left[ x_{1},x_{2},\ldots,x_{n}\right] $. Let
$G\in\mathcal{P}$ be nonzero. Then, there exist $n$ scalars $a_{1}
,a_{2},\ldots,a_{n}\in\mathbb{K}$ such that
\begin{equation}
G\left( a_{1},a_{2},\ldots,a_{n}\right) \neq0.
\label{darij1.eq.t3.1}
\tag{3}
\end{equation}
Theorem 2 is a particular case of the known fact that any polynomial ring over
an integral domain must itself be an integral domain. See, e.g.,
math.stackexchange question #2187381
for a proof in the case of univariate polynomial rings; but the general case
of multivariate polynomial rings can be reduced to this case by induction
(just adjoin the $n$ indeterminates one by one).
Theorem 3 is a well-known fact that is often stated (somewhat imprecisely, but
"morally right") in the form "$\mathbb{K}^{n}$ is Zariski-dense in
$\mathbb{K}^{n}$"; it is the reason why the old-fashioned habit of identifying
polynomials (defined formally as families of coefficients) with polynomial
functions (i.e., functions from $\mathbb{K}^{n}$ to $\mathbb{K}$ that are
given by a polynomial formula) is harmless (when $\mathbb{K}$ is infinite!).
(If Theorem 3 was false, then there would exist different polynomials that
give rise to the same polynomial function, and thus we could not identify the
former with the latter. This indeed happens when $\mathbb{K}$ is finite; for
example, the univariate polynomials $x^{2}-x$ and $0$ over $\mathbb{F}_{2}$
are distinct, but the corresponding functions from $\mathbb{F}_{2}$ to
$\mathbb{F}_{2}$ are identical.)
Since you seem to be interested in finite-dimensional structures, let me give
a nonstandard proof of Theorem 3 (more precisely, a reference):
Proof of Theorem 3. The polynomial $G$ has at least one nonzero coefficient
(since it is nonzero). Let us pick such a coefficient of largest possible
degree. Let this be the coefficient before $x_{1}^{t_{1}}x_{2}^{t_{2}}\cdots
x_{n}^{t_{n}}$. Then, $\deg G=t_{1}+t_{2}+\cdots+t_{n}$. Note that
$t_{1},t_{2},\ldots,t_{n}$ are finite numbers, while $\mathbb{K}$ is an
infinite field; thus, $\left\vert \mathbb{K}\right\vert >t_{i}$ for each
$i\in\left\{ 1,2,\ldots,n\right\} $. Hence, the Combinatorial
Nullstellensatz (Theorem 1.2 in Noga Alon, Combinatorial
Nullstellensatz)
(applied to $F=\mathbb{K}$ and $f=G$ and $S_{i}=\mathbb{K}$) yields that there
are $s_{1}\in\mathbb{K}$, $s_{2}\in\mathbb{K}$, $\ldots$, $s_{n}\in\mathbb{K}$
such that $G\left( s_{1},s_{2},\ldots,s_{n}\right) \neq0$. Consider these
$s_{1},s_{2},\ldots,s_{n}$. Hence, there exist $n$ scalars $a_{1},a_{2}
,\ldots,a_{n}\in\mathbb{K}$ such that $G\left( a_{1},a_{2},\ldots
,a_{n}\right) \neq0$ (namely, $a_{i}=s_{i}$). Thus, Theorem 3 is proven.
$\blacksquare$
Now Theorem 1 is a stone's throw away:
Proof of Theorem 1. Assume the contrary. Thus, $F\neq0$. Hence, we know that
the polynomials $F,R_{1},R_{2},\ldots,R_{m}$ are nonzero (since we already
know that $R_{1},R_{2},\ldots,R_{m}$ are nonzero). Thus, their product
$FR_{1}R_{2}\cdots R_{m}$ is nonzero as well (since Theorem 2 shows that
$\mathcal{P}$ is an integral domain). Hence, Theorem 3 (applied to
$G=FR_{1}R_{2}\cdots R_{m}$) shows that there exist $n$ scalars $a_{1}
,a_{2},\ldots,a_{n}\in\mathbb{K}$ such that
\begin{align*}
\left( FR_{1}R_{2}\cdots R_{m}\right) \left( a_{1},a_{2},\ldots
,a_{n}\right) \neq0.
\end{align*}
Consider these $a_{1},a_{2},\ldots,a_{n}$. Now,
\begin{align}
& F\left( a_{1},a_{2},\ldots,a_{n}\right) \cdot\prod_{i=1}^{m}R_{i}\left(
a_{1},a_{2},\ldots,a_{n}\right) \nonumber\\
& =\left( FR_{1}R_{2}\cdots R_{m}\right) \left( a_{1},a_{2},\ldots
,a_{n}\right) \neq0.
\label{darij1.pf.t1.0}
\tag{4}
\end{align}
But a product can only be nonzero if all its factors are nonzero. Thus,
\eqref{darij1.pf.t1.0} entails
\begin{equation}
F\left( a_{1},a_{2},\ldots,a_{n}\right) \neq0
\label{darij1.pf.t1.2}
\tag{5}
\end{equation}
and
\begin{equation}
R_{i}\left( a_{1},a_{2},\ldots,a_{n}\right) \neq0\qquad\text{for all }
i\in\left\{ 1,2,\ldots,m\right\} .
\label{darij1.pf.t1.3}
\tag{6}
\end{equation}
Thus, \eqref{darij1.eq.t1.2} shows that $F\left( a_{1},a_{2},\ldots
,a_{n}\right) =0$ (since we have \eqref{darij1.pf.t1.3}). But this
contradicts \eqref{darij1.pf.t1.2}. This contradiction shows that our
assumption was false. $\blacksquare$
Theorem 3 can be reframed as a linear-algebraic statement: Namely, the map
\begin{align*}
\mathcal{P} & \rightarrow\prod_{\left( a_{1},a_{2},\ldots,a_{n}\right)
\in\mathbb{K}^{n}}\mathbb{K},\\
F & \mapsto\left( F\left( a_{1},a_{2},\ldots,a_{n}\right) \right)
_{\left( a_{1},a_{2},\ldots,a_{n}\right) \in\mathbb{K}^{n}}
\end{align*}
that sends each polynomial $F\in\mathcal{P}$ to the family $\left( F\left(
a_{1},a_{2},\ldots,a_{n}\right) \right) _{\left( a_{1},a_{2},\ldots
,a_{n}\right) \in\mathbb{K}^{n}}$ of all its values at points in
$\mathbb{K}^{n}$ is a $\mathbb{K}$-linear map (and even a $\mathbb{K}$-algebra
homomorphism). Theorem 3 states that this $\mathbb{K}$-linear map is
injective. This is equivalent to saying that the images of the monomials in
$\mathcal{P}$ under this map are $\mathbb{K}$-linearly independent. This
viewpoint is occasionally useful, but (to my knowledge) not here. Note that we
are talking about an infinite family of monomials, but of course linear
independence can be rephrased in finitary terms (just show that finite
subfamilies are linearly independent).
