Is this a valid way to prove, that $a^2 + b^2 \neq 3c^2$ for all integers $a, b, c$ ? (except the trivial case)

Suppose a solution with at least one of $a,b\neq 0$ exists. If $(a,b,c)$ is a solution then $(\frac{a}{\gcd(a,b)},\frac{b}{\gcd(a,b)},\frac{c}{\gcd(a,b)})$ is also a solution (and all terms are integers), and $a$ and $b$ are coprime.

Now notice that $a^2+b^2$ cannot be a multiple of $3$ unless both are multiples of $3$ (because $a^2\equiv 0$ or $1\bmod 3$). We conclude that if $a^2+b^2=3c^2$ we have $a=b=0$.

There is truth in your method for case 8. It is called inifinte descent and equivalent to induction (i.e., alternatively you might start away with assuming that $(a,b,c)$ is the smallest non-trivial solution, and then $(a/2,b/2,c/2)$ cannot be a solution).