Show that if $\gcd (x, y) = 1$ and $x^2 + y^2 = z^2$ then $\gcd (x, z) = \gcd (y, z) = 1$.
Let $g=\gcd(x,z)$. This means, that for some integers $x',z'$ we have $x=gx'$ and $z=gz'$. And so
$$ x^2 + y^2 = z^2 \quad \implies \quad g^2(z'^2 - x'^2) = y^2 $$
This means $g$ is a factor of $y$.
But if $g$ is a factor of $y$ and $x$ then $g=1$ since we know $\gcd(x,y)=1$. And so $\gcd(x,z)=1$.
By a symmetric argument $\gcd(y,z)=1$.
And so, the constraints $\gcd (x, y) = 1$ and $x^2 + y^2 = z^2$ imply $\gcd (x, z) = \gcd (y, z) = 1$.