I've written down the following proof:
Given positive integers $x,y,z,k$ we have: $$\;\;\;\;x^2+y^2+z^2=k^2, \rightarrow x^2+y^2=(k-z)(k+z)$$ We now let $k=z+1$, thus: $$x^2+y^2=2z+1$$ Let $z$ be even i.e. $z=2z'$, then:$$\;\;\;\;x^2+y^2=4z'+1$$ We know that there exist an infinite amount of primes of the form $4m+1$, therefore there are infinitely many $z'$s such that $4z'+1$ is prime.
By Fermat's theorem in additive number theory we know that if $p \equiv 1 \pmod4$ then $p=u^2+v^2$ for some positive integer $u$ and $v$, since for every prime of the form $4z'+1 \equiv 1 \pmod4$, there are infinitely many solutions to $$ therefore there are infinitely many solutions to $$.
I know this proof is probably an "overkill" for the question and that I should prove the two statements I used (infinitely many primes and F.Theorem), but I think the proof is nonetheless correct right? Also, how can I prove that there are infinitely many solutions to $$ such that $(x,y,z)=1$ ?