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The discriminant of $y^2=x^3-9$ is $-16\cdot 27\cdot 81=-2^4{3}^7$, so five and seven are primes of good reduction of an elliptic curve. Count solutions over their residue fields.

Modulo five the values of $x^3-9$ for $x=0,1,2,3,4$ are $1,2,4,3,0$. They yield $2,0,2,0,1$ affine points, respectively, and the point at infinity gives $\#E({\mathbb{F}}_5)=6$.

Modulo seven the values for $x=0,\dots ,6$ are $5,6,6,4,6,4,4$. Only the three values four are nonzero squares, giving six affine points and $\#E({\mathbb{F}}_7)=7$.

For a rational elliptic curve at an odd prime of good reduction of an elliptic curve, the kernel of reduction on ${\mathbb{Q}}_p$ has no torsion element: it is the formal group on $p{\mathbb{Z}}_p$, whose formal logarithm embeds it in the additive group $p{\mathbb{Z}}_p$. Thus the entire torsion subgroup injects into each of the two finite reduction groups. Its order divides both six and seven, hence is one:Equivalently one may use prime-to-residue-characteristic injection at both primes; it gives the same coprime bound. This is a two-prime bound on rational elliptic torsion.

The elliptic-curve discriminant is $37$, so the elliptic curve has good reduction at both $2$ and $3$. At a prime $\ell$ of good reduction, the reduction of torsion points on an elliptic curve is injective on torsion of order prime to $\ell$. Here is the formal reason: the kernel of reduction of an elliptic curve is described by an integral formal group law on $\ell {\mathbb{Z}}_{\ell }$, and its multiplication series is $[n](t)=nt+O(t^2)$. If $\ell \nmid n$, the first term has strictly smaller valuation than every higher term for $t\ne 0$. Thus this kernel has no nonidentity point killed by $n$.

Apply this to each prime-primary component of the rational torsion subgroup. A component at $2$ injects into ${\tilde{E}}_3({\mathbb{F}}_3)$, of order $7$, and is therefore trivial. A component at $3$ injects into ${\tilde{E}}_2({\mathbb{F}}_2)$, of order $5$, and is trivial. Every other prime-primary component injects into both groups, so its order divides $\gcd (5,7)=1$. This proves the two-prime bound on rational elliptic torsion in the present case:

Use the two-prime bound on rational elliptic torsion with the counts six at three and eight at five from solution. For a prime $\ell \ne 3,5$, the $\ell$-primary torsion subgroup injects into both reduced groups by reduction of torsion points on an elliptic curve. Its order therefore divides the corresponding primary part of $\gcd (6,8)=2$. The three-primary subgroup injects into the group of order eight at five and is trivial; the five-primary subgroup injects into the group of order six at three and is trivial. Hence the entire rational torsion subgroup has order at most two.

The factorization $x^3+x^2+x+1=(x+1)(x^2+1)$ supplies $T=(-1,0)$, a nonidentity point of order two under the elliptic curve group law. ThereforeThe primary-component argument avoids assuming injectivity on residue-characteristic torsion.

Use reduction of torsion points on an elliptic curve, which is injective on torsion of order prime to the residue characteristic. Any two-primary torsion point of an elliptic curve must inject into the group of order seven at three, so is zero. Any three-primary torsion point of an elliptic curve injects into the group of order five at two, so is zero. For a prime $q\ne 2,3$, its primary torsion injects into both reduced groups and would have order dividing both five and seven, so is also zero. Decomposing the cyclic group generated by any torsion point of an elliptic curve into its primary parts shows that $E(\mathbb{Q})$ has no nonidentity torsion at all. This is the two-prime bound on rational elliptic torsion, with the residue-characteristic exceptions handled explicitly. Since $P=(0,0)$ is not $O$,