 Solution

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.