= Solution
Use <two-isogeny descent> with $a=-7$, $b=12$, so the partner <elliptic curve> is $E':Y^2=X^3+14X^2+X$. On $E$, a negative $x$ would make $x(x-3)(x-4)$ negative. Hence the <prime-support bound in two-isogeny descent> restricts the <square class> image to $\{1,2,3,6\}$. All four occur: the identity gives $1$, and the <rational points> $(2,2)$, $(0,0)$, $(6,6)$ give respectively $2$, $12\equiv3$, and $6$ in $\mathbb Q^*/\mathbb Q^{*2}$. Thus $|\alpha E(\mathbb Q)|=4$.
For the partner, $b'=1$, so its <square class> image is contained in $\{1,-1\}$. A realization of $-1$ would yield coprime integers $U,V$, not both zero, satisfying
$$
N^2=-U^4+14U^2V^2-V^4.
$$
If exactly one of $U,V$ is odd, the right side is $3$ modulo four, which is not a square. If both are odd, $U^4\equiv V^4\equiv1\pmod {16}$ and $14U^2V^2\equiv14\pmod {16}$, so the right side is $12$ modulo sixteen, again impossible. Both even is excluded by coprimality. This proves the <negative unit obstruction for the fourteen-coefficient isogeny cover>; the identity realizes $1$, so $|\alpha'E'(\mathbb Q)|=1$.
The <two-isogeny rank formula> now gives $2^r=4\cdot1/4=1$. Consequently this is the <rank-zero elliptic curve with roots zero three and four>:
$$
\boxed{\operatorname{rank}E(\mathbb Q)=0.}
$$
The displayed nonzero-ordinate points are therefore <torsion points of an elliptic curve>, rather than evidence of positive rank. In fact doubling either $(2,2)$ or $(6,6)$ gives $(4,0)$, so both have order four.
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