Show $x^2+y^2=9z+3$ has no integer solutions 
Show $x^2+y^2=9z+3$ has no integer solutions

So I know that $x^2+y^2=3(3z+1)$
And since $3\mid (9z+3)$ and $3\not\mid (3z+1)$ then $9z+3$ cannot be a square since it has a prime divisor which has a power less then $2$.
So does know this isn't a pythagorean triple $(x,y,\sqrt{9z+3})$ give me that there are no solutions or do I have to show more?

You've shown that $3|9x + 3$.  You've also shown that $9\not \mid 9x+3$.
SO that means $3|x^2 + y^2$ so but $9\not \mid x^2 + y^2$.
If $x$ and $y$ are divisible by $3$ then $x^2 + y^2$ is divisible by $9$.  So that's out.  If one of $x$ or $y$ is divisible by $3$ but the other isn't then $x^2 + y^2$ is not divisible by $3$ and that's out.
So the only options is neither $x$ nor $y$ is divisible by $3$.
So $x \equiv \pm 1 \pmod 3$ and $y \equiv \pm 1 \pmod 3$.  Which means $x^2 \equiv (\pm 1)^2 \equiv 1 \pmod 3$ and $y^2 \equiv (\pm 1)^2 \equiv 1 \pmod 3$.
So $x^2 + y^2 \equiv 2\pmod 3$ and $3\not \mid x^2 + y^2$ so that's out.
So all three options are out.
