$ 7\mid x \text{ and } 7\mid y \Longleftrightarrow 7\mid x^2+y^2 $
Solution 1:
Let $x,y \in \mathbb{F}_p$ be with $x^2+y^2=0$. If $x=0$, then $y=0$. Now assume $x \neq 0$. Let $z:=y/x$, then $z^2=-1$. If $p=2$, this means $z = 1$. If $p > 2$, this means that $z$ has order $4$ in $\mathbb{F}_p^*$, which happens iff $4|p-1$ i.e. $p \equiv 1 \bmod 4$. Hence, for every odd prime $p$ with $p \not\equiv 1 \bmod 4$ the quadratic form $x^2+y^2=0$ has only the trivial solution.
(While your method for $p=7$ is fine, try it with $p=67$!)