The pythagorean triples $x^2 + y^2 = z^2$ can be solved in integers using rational parameterization of solutions to $x^2 + y^2 = 1$.
It goes through $(1,0)$, then consider the line $y = -k (x - 1)$ so that $x^2 + k^2(x-1)^2 = 1$
We get $(1+ k^2 )x^2 - 2k^2 x + k^2 = 1$ or $x = \frac{k^2-1}{k^2+1}$ and $y=\frac{-2k}{k^2+1}$ and $z=1$
Then set $k= m/n$, for $(x,y,z) = (k^2-1,2k,k^2+1)=(m^2-n^2,2mn,m^2+n^2)$
What happens for 60-degree angle triangles $x^2+xy+y^2 = z^2$ ? We could look for rational solutions to
$x^2 + xy + y^2 = 1 $ and $(1,0)$ works again. Intersect with the line $y= k (x-1)$ ...
get integer solutions: $(m^2-n^2, -m^2+2mn, m^2 - m n + n^2)$ A similar derivation was obtained earlier on math.StackExchange
In the case of pythagorean triple we can build new solutions $(m^2-n^2,2mn,m^2+n^2)$ from old using the maps
$ (m,n) \mapsto (2m-n,m) \text{ or } (2m+n,m) \text{ or } (m+2n,n)$
Can I find something similar to the Pythagorean triple tree for this quadratic form, $x^2 + xy + y^2 = z^2$?