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I first heard of the ZF Aziom of Pairing watching this. I don't get why it is necessary to have an axiom which states that a set exists. Doesn't a set exist simply by virtue of the fact that it has a definition?
Given that $x$ and $y$ exist, why can I not claim that $S=\{x,y\}$ exists without this axiom?
I think that your view of set theory is not formal enough. Anyone will happily introduce set theory to a newcomer by describing sets as "collections of stuff". Then one might say that the formula $x\in A$ means that "the object $x$ is a member of the collection $A$", and that $\{a,b,c\}$ is a collection that contains exactly the objects $a$, $b$ and $c$. But this is just psychological.
Formally speaking, set theory deals with completely opaque objects, which we choose to call "sets", and which can be related by some binary relation $\in$, which we choose to call "membership". But just as Hilbert famously suggested to rename points, lines and planes in geometry as chairs, tables and beer glasses, we might rename "sets" as "giraffes" and "membership" as "bumbleness" (so a giraffe $y$ might bumble another giraffe $x$, which we will denote as $x\in y$).
Then Giraffe Theory explains how the bumbleness relation behaves and relates giraffes. We might ask: given two giraffes $x$ and $y$, is there a giraffe $A$ such that the complete list of giraffes which $A$ bumbles consists of exactly $x$ and $y$? Well, a priori there is no reason why that would be the case. I don't even really know what "bumble" means!
And then someone comes along and says: "Well of course there is one, I denote it by $\{x, y\}$! It exists because I can write that symbol." I would ask "But what does $\{x,y\}$ mean?" And that person would answer "It is precisely the giraffe which bumbles $x$ and $y$, and nothing else. By definition." And the obvious reaction would be "The fact that you can write this symbol and declare that it represents such a giraffe does not mean that this giraffe actually exists."
The point of this little charade is to make you realize that your mistake is to think that the set $\{x,y\}$ "obviously" exists, because you know what a set is supposed to be and how sets are supposed to behave. But the point of set theory is precisely to formalize that, so we may not assume anything. The reason why the pairing axiom is necessary is that otherwise we can't prove that there is such a pair set, using the other axioms. You may view the pairing axiom as the formal justification for which you are allowed to write the symbol $\{x,y\}$ (and more generally $\{x_1,\dots,x_n\}$ for a finite number of objects) and declare that this is a well-defined set which contains exactly $x$ and $y$.
Without being able to prove in ZFC that the unordered pair $\{x,y\}$ exists whenever $x, y$ are sets, all you have is a notation, $\{x,y\}$, for that unique set which, if it exists, contains just $x$ and $y$. The notion that "a set exist[s] simply by virtue of the fact that it has a definition" is codified by the (false, inconsistent) "naive Comprehension" principle, which claims that for any formula $\psi(x)$, the collection $\{x\,|\,\psi(x)\}$ is a set. Taking $\psi(x)$ to be $x\notin x$ gives the Russell paradox.
Pairing is redundant in the presence of other axioms. Given the Axiom of Infinity (in its usual formulation) and Separation, you can show that the set $2 = \{\emptyset, \{\emptyset\}\} = \{0, 1\}$ exists. Similarly, If you have existence of $\emptyset$ and the Powerset axiom, again $2 = \mathcal{P}(\mathcal{P}(\emptyset))$ is provably a set.
Now, given any sets $a, b$, the class function $F$ given by $0 \mapsto a, 1 \mapsto b$ is definable — for example, by this formula with parameters $a, b$:
$$\varphi(x, y) =_{def} (x=0\land y=a)\lor(x=1\land y=b).$$
Because $2$ as above is a set, by Replacement its image under $F$ is a set. That image is $F[2] = \{y\,|\, (\exists x\in 2)\, \varphi(x, y)\} = \{a, b\}$.
Pairing exists as a separate axiom because we want to be able to study the systems that result from dropping other ZFC axioms such as Infinity, Replacement, or Powerset. The remaining core of axioms should characterize basic aspects of "sethood" that don't involve those notions. Pairing is a basic requirement we have for universes of sets — whatever else a model of set theory does or doesn't contain, surely it ought to be closed under pairing.
Note that Separation can be proved from Replacement and the other ZFC axioms, but it's retained in most presentations as a separate axiom schema, for pedagogical reasons — it provides an opportunity to discuss naive Comprehension and the resulting inconsistencies — and because ZC, ZFC minus Replacement, came first and remains a thing.
"Doesn’t a set exist by virtue of it’s definition"
No, essentially you are saying that if you have a property $P$, then there should exist a set $X$ whose elements are exactly those which satisfy $P$.
Now this may seem like a very natural axiom, however it’s inconsistent(using it you can prove absurd statements like $0=1$!).
One can see it’s inconsistency by letting $P(x)=\neg(x\in x)$ and forming the “set of all sets which do not contain themselves” which leads to the famous Russel Paradox.
See https://en.wikipedia.org/wiki/Russell%27s_paradox
