Is every infinite set equipotent to a field?

The field axioms are a countable set over a countable language. Since there is an infinite model of the field axioms, there is (by Lowenheim-Skolem) a model of every infinite cardinality.

Is it true that for any infinite set $E$, the cardinal of the field $\mathbb Q(E)$ is equal to the cardinal of $E$?

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What is $\mathbb Q(E)$?
Here $E$ is considered to be a set of "indeterminates". Or, if that is not convenient, invent a set of indeterminates perhaps with some notation like $X_e$, one for each element $e \in E$. Then $\mathbb Q[E]$ is the set of polynomials in these indeterminates, with coefficients in $\mathbb Q$. (Of course each polynomial involves only finitely many of the elements of $E$.)

So $\mathbb Q(E)$ is a ring. It is an algebra over $\mathbb Q$. It is an integral domain.

And $\mathbb Q(E)$ is the set of rational functions in these indeterminates. In other words, formulas of the type $f/g$, where $f,g \in \mathbb Q[E]$ and $g \ne 0$.

So $\mathbb Q(E)$ is a field. It has the set $E$ (or a set $\{X_e : e \in E\}$ identified with $E$) of mutually algebraically independent elements that generate it (as a field over $\mathbb Q$).

To compute the cardinality of $\mathbb Q(E)$, compute in turn: how many monomials in $E$ are there; how many linear combinations of monomials (i.e. polynomials); how many quotients of those (rational functions).

For every infinite set $X$ there is a field $\mathbb F$ which is equipotent to $X$. In fact, much more is true. This is an immediate consequence of the Upward Löwenheim-Skolem Theorem. Fix a countable field $K$ (e.g. $\mathbb Q$). By the Löwenheim-Skolem Theorem there is some $\mathbb F$ such that $\mathbb F$ is equipotent to $X$ and $K \prec \mathbb F$. This implies that $K$ and $\mathbb F$ have the same theory and since $K$ is a field, $\mathbb F$ is a field as well.