x² + y² = 3z² has no non-trivial solution

Idea. Descent: from a non-zero solution build a strictly smaller one.

Proof. Suppose there is a solution other than $(0,0,0)$, and choose one with $|z|$ smallest positive (if $z=0$ then $x^2+y^2=0$ forces $x=y=0$).

Modulo $3$. $x^2+y^2\equiv0\pmod3$. Since squares are $0$ or $1$ modulo $3$, the sum is $0$ only if both are $0$, so $3\mid x$ and $3\mid y$. Write $x=3x_1$, $y=3y_1$:
$$9(x_1^2+y_1^2)=3z^2\ \Longrightarrow\ z^2=3(x_1^2+y_1^2),$$
so $3\mid z$; write $z=3z_1$. Then $9z_1^2=3(x_1^2+y_1^2)$, i.e.
$$x_1^2+y_1^2=3z_1^2,$$
a solution with $|z_1|=|z|/3<|z|$ and $z_1\ne0$. Contradiction with minimality.

Lesson. Infinite descent: reduce modulo a well-chosen number, show all variables are divisible by it, divide, and get a smaller solution.