The main proof you mention is the easiest and the best since it generalizes very well. If you really want something more intrinsic, note
$$(a+b\sqrt 2)(c+d\sqrt 2)=0\implies (a^2-2b^2)(c^2-2d^2)=0$$
But then if so, either $a^2=2b^2$ or $c^2=2d^2$, WLOG assume the former.
Then $a$ is even, but then if the prime factorization of $a$ is $2^kp_1^{e_1}\ldots p_r^{e_r}$ we have
$$a^2=2^{2k}p_1^{2e_1}\ldots p_r^{2e_r}$$
So the exponent of $2$ in $a$ is even, however since $b^2$ has an even power of $2$, $2b^2$ has an odd power, a contradiction, unless that power is infinity, but the only number infinitely divisible by $2$ is $0$, so $a=b=0$ holds