The only thing I don't [understand] is the last part where it's given as "One may thus equate numerators with numerators and denominators with denominators, giving Euclid's formula"
It uses unique fractionization $\Rightarrow$ uniqueness of reduced fractions (with denominators $> 0),\,$ i.e.
$\qquad\qquad \begin{align}\gcd(\color{#c00}{c,b})=1\\ \gcd(j,k)= 1\end{align}$, $\ \ \dfrac{c}b = \dfrac{j}k\ \Rightarrow\ \begin{align} c&\,=\,j\\ b &\,=\, k\end{align},\ \ \ {\rm for}\ \ b,c,j,k\in \Bbb Z,\ b,k > 0$
Follow the link for a simple proof using Euclid's Lemma (hint: $\,j = ck/b\,\Rightarrow\,\color{#c00}{b\mid c}\,k\,\Rightarrow\,b\mid k)$
Remark $ $ A more conceptual way to derive this parametrization of Pythagorean triples is to employ arithmetic of Gaussian integers $\,\Bbb Z[i] = \{ a + b\,i\,: a,b\in\Bbb Z\}$. Like integers they enjoy (Euclidean) division (with smaller remainder) and this implies they too satisfy the analog of the Fundamental Theorem of Arithmetic = existence and uniqueness of factorization into primes (= irreducibles). This implies that coprime factors of a square must themselves be squares (up to unit factors $\,\pm1,\pm i)$
Thus if $\ z^2 = x^2 + y^2 = (x-y\,i) (x+ y\,i) $ and $\,x,y\,$ are coprime then one easily checks that $\,x-y\,i,\,x+y\,i\,$ are coprime, so being coprime factors of the square $\,z^2$ they must themselves be squares (up to a unit factor). Thus e.g. $\ x + y\ i\, =\, (m + n\ i)^2 =\ m^2 - n^2 + 2mn\, i,\,$ hence $\,x = m^2-n^2,\ y = 2mn\,$ (using the unit factor $1$; using the other unit factors $\, -1,\pm i\,$ merely changes signs or swaps $\,x,y\,$ values). Notice how very simple the solution is from this perspective.
This is a simple prototypical (arithmetical) example of the simplification that results from the transformation of nonlinear problems into linear problems by working in larger algebraic extension rings. See here for some further discussion of such.
Best Answer
As stated in the Wikipedia article, the set of ALL pythagorean triples $(a,b,c)$ is given by: $$ a = k(m^2 - n^2) \;;\quad b = k(2mn) \;;\quad c = k(m^2 + n^2) \tag{1} $$ where $m, n, k$ range over the positive integers, $m - n$ is odd, $m > n$, and $m,n$ are relatively prime. You can also switch $a$ and $b$ above, if you like, to get all triples where order of (a,b) matters. So anyway, to answer your questions:
Yes, there are infinitely many pythagorean triples. The easy way to show this is to take one triple, say $3, 4, 5$, and take all multiples of it. That corresponds to letting $k$ range over all integers in (1). But there infinitely many primitive triples, too (ones that aren't just multiples of a smaller triple); this is because there are infinitely many pairs $m, n$ with $m - n$ odd, $m > n$, and $m, n$ relatively prime.
For any integer multiple of four $l$, you can certainly write it as $l = 2mn$ with $m,n$ relatively prime, $m - n$ odd. For an odd integer $l \ge 3$, note that it is the difference between consecutive squares, so take $m = n+1$, where $l = (n+1)^2 - n^2 = 2n+1$. For an even integer $l \ge 6$ that is not a multiple of four, it won't be part of a primitive triple, but it will be part of a triple--just find a triple for $\frac{l}{2}$, and then multiply each term by $2$.
In summary:
There are infinitely many pythagorean triples.
There are also infinitely many primitive pythagorean triples.
Every positive integer $\ge 3$ is part of a pythagorean triple.
Every positive integer $\ge 3$ that is not congruent to $2$ mod $4$ is part of a primitive pythagorean triple.
$1$ and $2$ are not part of any pythagorean triples, though they would be part of $0, 1, 1$ and $0, 2, 2$ if we allowed these trivial cases.
P.S. You may also be interested in the infinite tree of primitive pythagorean triples.