Question 1: Putting both curves in say, Legendre Normal Form (or else appealing the lefschetz principle) shows that if the two curves are isomorphic over $\mathbf{C}$ then they are isomorphic over $\overline{\mathbf{Q}}$. Now we could say that for instance $E_2$ is an element of $H^1(G_{\overline{Q}}, Isom(E_1))$ where we let $Isom(E_1)$ be the group of isomorphisms of $E_1$ as a curve over $\mathbf{Q}$ (as in Silverman, to distinguish from $Aut(E_1)$, the automorphisms of $E_1$ as an Elliptic Curve over $\mathbf{Q}$, that is, automorphisms fixing the identity point). However, $E_2$ is also a principal homogeneous space for a unique curve over $\mathbf{Q}$ with a rational point, which of course has to be $E_2$, so the cocycle $E_2$ represents could be taken to have values in $Aut(E_1)$. Now $Aut(E_1)$ is well known to be of order 6,4 or 2 depending on whether the $j$-invariant of $E_1$ is 0, 1728 or anything else, respectively. Moreover the order of the cocycle representing $E_2$ (which we now see must divide 2, 4 or 6) must be the order of the minimal field extension $K$ over which $E_1$ is isomorphic to $E_2$. So $K$ must be degree 2,3,4 or 6 unless I've made an error somewhere.
Question 2: If you restrict your focus to just elliptic curves, yes your idea is right. If it's a quadratic extension, you have exactly 1 non-isomorphic companion. If you have a higher degree number field, you have nothing but composites of the quadratic case unless your elliptic curve has j invariant 0 or 1728.
Notice I am very explicitly using your choice of the word elliptic curve for both of these answers.
Theorem (originally due to Setzer?): Fix $E/\mathbb{Q}$ with $j(E)$ not 0 or 1728. Then for all but finitely many inequivalent twists $E_d$, the torsion subgroup $E_d(\mathbb{Q})_{tors}$ is isomorphic to $E[2](\mathbb{Q})$, so in particular $E_d(\mathbb{Q})_{tors}$ has order 1, 2, or 4. (Probably he also proved it for number fields.)
There's a paper of mine$^1$ with a much more general theorem using the theory of heights. I don't recall Setzer's proof except that it doesn't use heights.
Theorem: Let $K$ be a number field and let $A/K$ be an abelian variety with $\mu_n\subset {\rm Aut}(A)$. (This means we can twist $A$ by $n$'th roots of $d$.) Then every point $P\in A_d(K)$ satisfies one of the following two conditions:
- $P$ is fixed by a non-trivial $\zeta\in\mu_n$.
- $\hat h(P) \ge C_1(A)h^{(n)}(d) - C_2(A)$.
Here $\hat h$ is the canonical height relative to an ample symmetric divisor, and $h^{(n)}(d)$ is a sort of "$n$'th power free height," say equal to the minimum of $h(du^n)$ for $u\in K^*$. The constants depend on $A$, but are independent of $d$.
It follows from the theorem that after discarding finitely many $d \in K^*/{K^*}^n$, a point in $A_d(K)$ is either $1-\zeta$ torsion (hence $nP=O$), or its canonical height is positive, and hence it is nontorsion.
Of course, to describe more precisely what happens for the finitely many exceptional $d$ can be a delicate matter, as some of the other answers have indicated. I think it's interesting to see how one can approach the problem via heights or via representation theory.
$^1$ J.H. Silverman, Lower bounds for height functions, Duke Math. J. 51 (1984), 395-403.
EDIT: Fixed statement of first theorem. I'd originally written that "for all but finitely many inequivalent twists $E_d$, the torsion subgroup $E_d(\mathbb{Q})_{tors}$ has at most two elements." This is clearly false, since if $E$ has the form $E:y^2=(x-a)(x-b)(x-c)$ with $a,b,c\in\mathbb{Q}$, then $E_{d}[2](\mathbb{Q})$ has order 4 for every twist.
Best Answer
There is a formula but it involves both cubic twists. Let $E: y^2 = x^3+B$ be an elliptic curve over $\mathbb{Q}$ with $j=0$ as the one in the question. Let $D$ be a cubefree integer. Set $E_1: y^2=x^3+D^2\,B$ and $E_2:y^2=x^3+D^4\,B$. Both $E_1$ and $E_2$ become isomorphic to $E$ over $L=\mathbb{Q}\bigl(\sqrt[3]{D}\bigr)$. I believe the formula is
$\DeclareMathOperator{\rk}{rk}$ $\rk E(L) = \rk E(\mathbb{Q}) + \rk E_1(\mathbb{Q}) + \rk E_2(\mathbb{Q}).$
This can easily checked to be the case for the given curve and some small $D$.
Let $K=L(\mu_3)$ be the Galois closure of $L$ and let $F=\mathbb{Q}(\mu_3)$. Denote by $G$ the cyclic Galois group of $K/F$ and write $\chi$ and $\bar \chi$ for the two non-trivial characters of $G$.
First the motivation: The $L$-function associated to the $G_F$-representation $T_p E\otimes \mathbb{C}[G]$ is the $L$-function of $E/K$. Let $\psi$ be the Grössencharacter associated to $E$, then this $L$-function splits into six $L$-functions, namely those for $\psi$, $\bar\psi$, $\psi\chi$, $\overline{\psi\chi}$, $\psi\bar{\chi}$, and $\bar{\psi}\chi$. The first two give the $L$-function of $E/F$, the middle two the $L$-function of $E_1/F$ and the last two the $L$-function of $E_2/F$. The Birch and Swinnerton-Dyer conjecture now implies that $\rk E(K) = \rk E(F)+\rk E_1(F)+\rk E_2(F)$.
We can prove this directly by looking at the $G$-action on $E(K)$. Write $g$ for the element of $G$ that sends $\alpha\in L$ with $\alpha^3 =D$ to $\zeta\cdot \alpha$ where $\zeta^3=1$. Let $[\zeta]\in \operatorname{End}(E)$ be the element of order $3$ given by $[\zeta](x,y) = (\zeta \cdot x, y)$. Define $E(K)_i = \bigl\{ P \in E(K)\, :\, g(P) = [\zeta]^i(P)\bigr\}$. Then $E(K)_0 = E(F)$ and the map $E_1(F) \to E(K)$ sending $(x,y)$ to $(x/\alpha^2, y/\alpha^3)$ has image equal to $E(K)_1$. Similar $(x,y)\mapsto (x/\alpha^4,y/\alpha^6)$ brings $E_2(F)$ to $E(K)_2$.
The kernel of the map from $E(F)\oplus E_1(F)\oplus E_2(F)$ to $E(K)$ is contained in the finite subgroup $E(F)[\zeta-1]$. For any $P\in E(K)$ we have $3P = (1+g+g^2)(P) + (1+[\zeta]g+[\zeta]^2g^2)(P) + (1+[\zeta]^2g+[\zeta]^4g^2)(P)$ which belongs to the sum $E(F)+E(K)_1+E(K)_2$. Thherefore the cokernel is also finite. Hence $\rk E(K) = \rk E(F)+\rk E_1(F)+\rk E_2(F)$.
Since $E$ has complex multiplication by an order in $F$, the rank of $E(F)$ is twice the rank of $E(\mathbb{Q})$; and similarly for $E_1$ and $E_2$. The representation $E(K)\otimes\mathbb{C}$ of the Galois group (equal to $S_3$) of $K/\mathbb{Q}$ splits as $\mathbb{1}^a + \epsilon^a+\rho^b$ where $\epsilon$ is the irreducible non-trivial $1$-dimensional and $\rho$ is the irreducible $2$-dimensional representation of $S_3$. Therefore $\rk E(L) = a+b = \tfrac{1}{2} (2a+2b) = \tfrac{1}{2} \rk E(K) = \tfrac{1}{2}\cdot \bigl( \rk E(F) + \rk E_1(F) + \rk E_2(F) \bigr) =\rk E(\mathbb{Q}) + \rk E_1(\mathbb{Q}) + \rk E_2(\mathbb{Q})$.