Legendary.....?????

First notice that the $R.H.S$ is $\equiv 3\pmod{4}$ and we know that a square is either $\equiv 0$ or $1\pmod{4}.$ $0$ when the square is even i.e the number is even and $1$ when the square is odd i.e the number is odd.This implies that $x,y,z$ are odd numbers.Let us say that $x=2a+1,y=2b+1,z=2c+1.$ for some $a,b,c\in\mathbb{Z}^+$ (they can also be equal to $0$ ).Substituting and simplifying a little gives, $a^2+a+b^2+b+c^2+c=\dfrac{340}{4}=85\\ \rightarrow a(a+1)+b(b+1)+c(c+1)=85$ .But,this isn't possible as the product of two consecutive integers is always even and thus, the $L.H.S$ would be even but the $R.H.S$ isn't.Thus, $0$ solutions.

Note $x^2 \equiv 0, 1, 4$ $(\mod 8)$ .
$343 \equiv 7$ $(\mod 8)$ .

No combinations of 3 numbers from $0, 1, 4$ add to make $7$ $(\mod 8)$ . Hence the answer is $0$ .

Legendres three square theorem See the question title.