Proving that $a^2+b^2\neq 3c^2$ for any positive nonzero integers $a$, $b$, and $c$

You proved correctly that if $a,b,c\in\mathbb N$ are such that $a^2+b^2=3c^2$, then $a$, $b$ and $c$ must all be even numbers. So, you have $a=2m$, $b=2n$, and $c=2p$. But then $a^2+b^2=3c^2$ means that $m^2+n^2=3p^2$. And now you can start all over again. This is impossible, since you can't divide a positive integer by $2$ again and again and to keep always getting positive integers.

Hint If $a,b,c$ all all even, cancel a 2 and repeat.