How many integer-sided right triangles are there whose sides are combinations? 
How many integer-sided right triangles exist whose sides are combinations of the form $\displaystyle \binom{x}{2},\displaystyle \binom{y}{2},\displaystyle \binom{z}{2}$?

Attempt:
This seems like a hard question, since I can't even think of one example to this. Mathematically we have,
$$\left(\dfrac{x(x-1)}{2} \right)^2+\left (\dfrac{y(y-1)}{2} \right)^2 = \left(\dfrac{z(z-1)}{2} \right)^2\tag1$$ 
where we have to find all positive integer solutions $(x,y,z)$. 
I find this hard to do. But here was my idea. Since we have $x^2(x-1)^2+y^2(y-1)^2 = z^2(z-1)^2$, we can try doing $x = y+1$. If we can prove there are infinitely many solutions to,
$$(y+1)^2y^2+y^2(y-1)^2 = z^2(z-1)^2\tag2$$ 
then we are done.
 A: Hint: Start the other way around, by the formula to generate all Pythagorean triples.
A: Solving $(1)$ for $z$, we have,
$$z = \frac{1\pm\sqrt{1\pm4w}}{2}\tag3$$
where,
$$w^2 = (x^2-x)^2+(y^2-y)^2\tag4$$
It can be shown that $(4)$ has infinitely many integer solutions. (Update: Also proven by Sierpinski in 1961. See link given by MXYMXY, Pythagorean Triples and Triangular Numbers by Ballew and Weger, 1979.)
However, the problem is you still have to solve $(3)$. I found with a computer search that with $x<y<1000$, the only integers are $x,y,z = 133,\,144,\,165$, so,
$$\left(\dfrac{133(133-1)}{2} \right)^2+\left (\dfrac{144(144-1)}{2} \right)^2 = \left(\dfrac{165(165-1)}{2} \right)^2$$
P.S. If you're curious about rational solutions, then your $(1)$ and $(2)$ have infinitely many.
A: For your equation, solutions include
$$(1,1,1)\quad (1,2,2)\quad (1,3,3)\quad (1,3,3)\quad
 (1,4,4)\quad (1,5,5)\quad \cdots $$
Any triple with zero or one for one of $\space x,y\space $ will work. All others would probably require a brute force search.
