It is easy to see that if $(z,u,w)$ is a primitive pythagorean triple, then $w$ is odd and either $z$ is odd and $u$ is even or $z$ is even and $u$ is odd. WLOG suppose that $u=2x$ is even, so $z$ is odd. Then
$$u^2=4x^2=w^2-z^2=(w+z)(w-z).$$
It is plain that $\gcd(w+z,w-z)=2$, so $y_1:=(w+z)/2$ and $y_2:=(w-z)/2$ are relatively prime positive integers. Thus
$$x^2=y_1y_2,$$
and therefore $y_1$ and $y_2$ are relatively prime divisors of $x^2$. This implies that $y_1$ and $y_2$ are perfect squares, so write $y_1=a^2$ and $y_2=b^2$. Hence,
$$u=2x=2ab,\quad z=y_1-y_2=a^2-b^2,\qquad\text{and}\qquad w=y_1+y_2=a^2+b^2.$$
This proves that every primitive triple has the form $(a^2-b^2,2ab,a^2+b^2)$, as wanted.
Of course, the condition $\gcd(a,b)=1$ is needed, otherwise $(a^2-b^2,2ab,a^2+b^2)$ won't be primitive.
If you can find one solution, $(x,y,z)$, then the answer to the question is "infinitely many," because for any positive integer $a$, $(ax,ay,az)$ will be another.
So to find one solution note that $2021 = 43\cdot 47.$ The equation can be rewritten:
$$43\cdot 47y^2 = z^2-x^2 = (z-x)(z+x).$$
$43$ and $47$ divide the left side, so they must divide the right. We guess that that maybe $z-x=43$ and $z+x = 47.$ This gives $z=45$, $x=2$, forcing $y=1$. So there's the one solution. You can probably construct others similarly.
Best Answer
If $n$ is odd, we have $$n^2 + \left(\dfrac{n^2-1}{2}\right)^2 = \left(\dfrac{n^2+1}{2}\right)^2$$ If $n$ is even we have $$n^2 + \left( \dfrac{n^2}{4}-1 \right)^2 = \left( \dfrac{n^2}{4} + 1 \right)^2$$