Clearly, we have solutions for any value of $z$, if we also have $x = y = 0$. We claim that there are no other solutions.

In order to show this, we first assume that $z$ is even. In this case, the number $7^z y^2$ is a perfect square, and therefore $x^4 + x^2 = x^2(x^2 + 1)$ must also be a perfect square. The only value of $x$ for which both $x^2$ and $x^2 + 1$ can be perfect squares is $0$, and we therefore have $x = 0$ (and thus also $y = 0$) in this case.

Now, assume that $z$ is positive and odd. Let $z = 2c + 1$. Since $x^2(x^2 + 1)$ is divisible by $7$ and $x^2 + 1$ can only be congruent to $1$, $2$, $3$ or $5$ modulo $7$, it follows that $x$ must be divisible by $7$. Let $x = 7^a u$ and $y = 7^b v$ with $u$ and $v$ not divisible by $7$. The given equation can now be written as
$$7^{2a} u^2 (7^{2a} u^2 + 1) = 7^{2(b + c) + 1} v^2,$$
which yields a contradiction, since the left-hand side of this expression is exactly divisible by an even number of sevens, while the right-hand side is exactly divisible by an odd number of sevens.

Finally, assume that $z$ is negative and let $z = -w$. In this case the equation can be written in the equivalent form $7^w x^2 (x^2 + 1) = y^2$. If $w$ is even, we once again note that both $x^2$ and $x^2 + 1$ must be perfect squares, and as before this means that $x = y = 0$ must follow. If $w$ is odd, we can write $w = 2c + 1$, and the same argument as before will hold, with $x = 7^a u$ and $y = 7^b v$ yielding $7^{2(a + c) + 1} u^2 (7^{2a} u^2 + 1) = 7^{2b} v^2$ as a contradiction.