This is in response to the comment-question of how to show:
if $F$ is finite, then $F[x]$ is countable
Notice that an element of $F[x]$ has finite length. Let us pick an $f \in F[x]$ and represent it as the function defined by:
$$x \mapsto a_0 + a_1 x^1 + a_2 x^2 + a_3 x^3$$
You can have more terms, but we'll use this just for the idea of the proof.
In the polynomial written out above, each $a_k \in F$, which means there are countably many of them. (If $F$ is finite, then there are finitely many; but you have also asked about the case in which $F = \mathbb{Q}$, which is countably infinite.)
Since $F$ is countable, there exists a map $g: F \rightarrow \mathbb{N}$ that takes elements of $F$ as input and gives natural numbers as outputs.
The idea here will be to consider the terms in order, beginning with the constant term and moving upwards in degree of the terms, and associate them with the prime numbers $2,3,5, \ldots$.
In particular, let us associate with the polynomial written above the natural number:
$$2^{g(a_0)}\cdot3^{g(a_1)}\cdot5^{g(a_2)}\cdot7^{g(a_3)}$$
This gives an injective (aka one-to-one) function from $F[x] \rightarrow \mathbb{N}$, which means that $F[x]$ is countable.
To write out the more general definition of our map, we just need to look at the example and figure out how to write it out in sum and product notation:
For any element of $F[x]$ written as
$$f := \sum_{n=0}^{k-1}a_{n}x^{n}$$
we write $P(k)$ to mean the $k^\text{th}$ prime and map $f$ to
$$\prod_{n=1}^{k}P(n)^{g(a_n)}$$
Since we have an injective function from $F[x] \rightarrow \mathbb{N}$, the claim of countability follows.
