Find rational solutions to $x^2 + y^2 = 6$ This question comes from Rational Points on Elliptic Curves (Silverman & Tate) Exercise $1.7$ (a).

Find rational solutions (if any) to $x^2 + y^2 = 6$

I think there exist no solutions and here is my proof:
Suppose that there is a rational point and write it as
$$x=\frac{X}{Z} \quad \text { and } \quad y=\frac{Y}{Z}$$
for some integers $X, Y$, and $Z$. Then
$$X^2 + Y^2 = 6Z^2$$
If $X, Y, Z$ have a common factor, then we may remove it, so we may assume that they have no common factor. It follows that neither $X$ nor $Y$ is divisible by $6$. This is true because if $6$ were to divide $X$, then $6$ divides $Y^2 = 6Z^2 - X^2$, so $6$ divides $Y$. But then $36$ divides $X^2 + Y^2 = 6Z^2$, so $6$ divides $Z$, contradicting the fact that $X, Y, Z$ have no common factors. Hence $6$ does not divide $X$, and a similar argument shows that $6$ does not divide $Y$.
Since $X$ and $Y$ are not divisible by $6$, we have
$$X \equiv \pm 1,4 \quad(\bmod 6) \text { and } Y \equiv \pm 1,4 \quad(\bmod 6),$$
and hence
$$X^2 + Y^2 \equiv 1+1 \equiv 2 \quad(\bmod 6)$$
or
$$X^2 + Y^2 \equiv 1+4 \equiv 5 \quad(\bmod 6)$$
or
$$X^2 + Y^2 \equiv 4+4 \equiv 3 \quad(\bmod 6)$$
However, we also have
$$X^2 + Y^2 = 3Z^2 \equiv 0 \quad(\bmod 6),$$
a contradiction.
I think this proof is sufficient. I based it on the method that the book showed there didn't exist any rational points on the curve $x^2 + y^2 = 3$.

The ideas in the proof are correct, but there are some minor errors that need correction to make your reasoning rigorous.
First, while it is true that $X,Z$ and $Y,Z$ may be taken to be relatively prime without loss of generality, and that this directly follows from $x = X/Z$ and $y = Y/Z$, it is not immediately obvious why $X,Y$ need to also be relatively prime.  To see why, consider the alternative equation $$X^2 + Y^2 = 40Z^2,$$ which of course admits integer solutions such as $(X,Y,Z) = (2,6,1)$.  Then $X, Z$ and $Y, Z$ are relatively prime but $X, Y$ are not.  So it is misleading to start off with such a claim; instead, it is better to omit it and directly proceed to your reasoning that neither $X$ nor $Y$ can be divisible by $6$.
Second, you have a few typographical errors:
$$X^2 + Y^2 \equiv 4 + 4 = 8 \equiv \color{red}{2} \pmod 6$$
and your last claim should read

However, we also have $$X^2 + Y^2 = \color{red}{6}Z^2 \equiv 0 \pmod 6.$$

