AustriaMO2011

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.