 Solution

Fix a prime $p$. If a rational point $T=(x,y)$ has $v_p(x)<0$, the integral projective Weierstrass equation forcesfor 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 thereforeFor $v_2(t)\ge 1$, 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.