<s>[INST] Show that there are no rational solutions on the ellipse $2x^2 + 3y^2 = 1$ Show that there are no rational solutions on the ellipse $2x^2 + 3y^2 = 1$.
Want to make sure that my proof is correct.
Suppose there are rational solutions.
Note $2x^2 + 3y^2 = 1 \iff 2x^2 + 3y^2 = z^2\ |\ x,y.z \in \mathbb{Z}$
Since taking $(\frac{x}{z}, \frac{y}{z})$ as our rational solution, $2(\frac{x}{z})^2 + 3(\frac{y}{z})^2 = 1 \implies 2x^2 + 3y^2 = z^2$.
Now reducing $mod\ 3$, we have $2x^2 + 3y^2 \equiv  1 \ mod\ 3 \implies 2x^2 \equiv 1 \ mod\ 3 \implies -1x^2 \equiv 1\ mod\ 3 \implies x^2 \equiv -1\ mod\ 3 $, which is impossible.
 [/INST] You correctly arrived at the following: If there are rational points on this ellipse then the equation$$2x^2+3y^2=z^2\tag{1}$$
has integer solutions $\ne(0,0,0)$. But starting from here the argument is more intricate than you assumed.
If there are integer solutions of $(1)$ then  after dividing them by a suitable power of $3$ we may assume that not all three of $x$, $y$, $z$ are divisible by $3$. Modulo $3$ we can rewrite $(1)$ as
$$z^2+x^2\equiv0\qquad{\rm mod}\ 3\ .$$
Since all squares are $=0$ or $=1$ modulo $3$ this is only possible if both $x$ and $z$ are $\equiv0$ modulo $3$. But then $(1)$ would imply that $3y^2=z^2-2x^2$ is divisible by $9$, hence $y$ would be divisible by $3$ as well, which we have excluded.
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