Baltic Way 2019

Assume that there exist positive integers $x$, $y$, $z$ satisfying the given equation. Let's consider two cases:

If $x$ is odd then $7^x\equiv 7(\mathrm{mod}8)$. Since each of $y^2$ and $z^2$ can have $0$, $1$, $4$ as the remainders when divided by $8$. We have contradiction because $1+y^2+z^2\equiv 1,2,3,5,6(\mathrm{mod}8)$.

If $x$ is even then $x=2n$. We have $$y^2+z^2=7^{2n}-1=(7^n+1)(7^n-1)$$ We can show that the right hand side has at least one prime factor $p\equiv 3(\mathrm{mod}4)$ of odd order. Thus $p|(y^2+z^2)$. It implies that $p|y$ and $p|z$. We have a contradiction because $p$ is a prime factor of even order of the left hand side.