I can't really seem to find an answer to this online. From what I understand, $E$ is an extension field of $F$ if $F \subseteq E$ and the operations of $F$ are those of $E$ restricted to $F$.
What I'm wondering is what the "largest" extension field can be. Are we going to reach a point where $E$ = $E'$, and there can be no extension field bigger than $E'$? Or, can we always constuct a larger extension field? Thank you.
As Hayden pointed out in the comments, given any field $F$, you can construct a strictly larger extension field by adjoining an indeterminate $x$. The field $F(x)$ is a transcendental extension of $F$ and is called the field of rational functions in one variable over $F$. You can thus produce the following infinite sequence of field extensions $$F \subsetneq F(x_1) \subsetneq F(x_1,x_2) \subsetneq F(x_1,x_2,x_3) \subsetneq\cdots.$$
If you modify your question by adding the hypothesis that the extension $F \supset E$ is algebraic, the answer changes. An extension $K/F$ is called algebraic if every element $\alpha \in K$ is the root of some nonzero polynomial $f(x) \in F[x]$. One can construct a special field called the algebraic closure of $F$, denoted by $\overline{F}$. There are multiple equivalent ways of defining the algebraic closure, but one way is to say an extension $K/F$ is an algebraic closure of $F$ if there are no nontrivial algebraic extensions of $K$, i.e. if $L/K$ is an algebraic extension, then $L$ is isomorphic to $K$.
As an example, the field of complex numbers $\mathbb{C}$ is an algebraic closure of the field of real numbers $\mathbb{R}$. We know this thanks to the fundamental theorem of algebra, which says that any polynomial $f(x) \in \mathbb{C}[x]$ has a root in $\mathbb{C}$ (this is another equivalent way of defining the algebraic closure).
So the moral of the story is that while there is no largest extension of $F$, there is a "largest" algebraic extension of $F$.