I want to find the abelian group of rational points $E(\mathbb{Q})_{\text{torsion}}$ of the elliptic curve $y^2=x^3-2$.
$$E(\mathbb{Q})_{\text{torsion}}=\{P \in E(\mathbb{Q}) | P \text{ of finite order }\}$$
From Lutz-Nagell theorem we have the following:
Let $E|_{\mathbb{Q}} , y^2=x^3+ax+b, a, b \in \mathbb{Z}$ and $P=(x, y) \in E(\mathbb{Q})$.
We suppose that $P$ is of finite order.
Then $x, y \in \mathbb{Z}$ and $y=0$ (that corresponds to the points of order $2$ ) or $y^2 \mid D(f)=4a^3+27b^2$.
We have also the following:
If $P(x, y) \in E(\mathbb{Q})$ and $x \notin \mathbb{Z}$ or $y \notin \mathbb{Z}$, then $P$ has infinite order.
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So, to find the points $P$ of finite order of the curve $y^2=x^3-2$ we do the following:
For $y=0 \Rightarrow x^3-2=0 \Rightarrow x \notin \mathbb{Z}$ (Do we have to say that $x \notin \mathbb{Z}$ or $x \notin \mathbb{Q}$ ?? )
so there is no rational point of order $2$.
Let $f(x)=x^3-2$, then $D(f)=27 \cdot 4=108$.
$$y^2 \mid D(f) \Rightarrow y^2 \mid 2^2 \cdot 3^3 \Rightarrow y^2 \mid 2^2 \cdot 3^2 \Rightarrow y \mid 6 \Rightarrow y=\pm 1, y =\pm 2, y=\pm 3 , y=\pm 6$$
For $y=\pm 1 \Rightarrow x^3-3=0 \Rightarrow x \notin \mathbb{Z}$.
For $y=\pm 2 \Rightarrow x^3-6=0 \Rightarrow x \notin \mathbb{Z}$.
For $y=\pm 3 \Rightarrow x^3-11=0 \Rightarrow x \notin \mathbb{Z}$.
For $y=\pm 6 \Rightarrow x^3-38=0 \Rightarrow x \notin \mathbb{Z}$.
So there are no rational points of the elliptic curve with finite order.
Is this correct??
Yes, it is correct that there are no rational points of finite order. And your application of the Nagell-Lutz Theorem seems valid too.