= Solution
Fix a prime $p$. If a rational point $T=(x,y)$ has $v_p(x)<0$, the integral projective Weierstrass equation forces
$$
v_p(x)=-2m,
\qquad v_p(y)=-3m
$$
for some $m>0$. Hence its formal parameter $t=-x/y$ lies in $p\mathbb Z_p$, so $T$ belongs to the <formal group of an elliptic curve> at the identity. This formal kernel has no nonzero torsion for the present equation. Multiplication by an integer prime to $p$ is a formal-group automorphism, while the <formal logarithm> rules out $p$-power torsion for odd $p$. For $p=2$, inversion sends $t$ to $-t$ because the equation has no $xy$ or $y$ term, and therefore
$$
[2]_F(t)=2t+O(t^3).
$$
For $v_2(t)\geq1$, its leading term has strictly smaller valuation than every higher term, so $[2]_F(t)$ cannot vanish; iterating excludes all two-power torsion as well.
Thus a nonzero torsion point cannot have $v_p(x)<0$. Since this holds for every prime, $x\in\mathbb Z$. The integral equation then makes $y^2$ an integer; a rational number whose square is integral is itself integral, so $y\in\mathbb Z$. This is the integrality assertion in the <Lutz–Nagell theorem>.
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