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.
If $p\ge5$ then $E$ has equation $y^2=x^3+Ax+B$
with $p\mid A$ and $p\mid B$. A quadratic twist alters
the discriminant, essentially $4A^3+27B^2$, by a sixth power,
so for it to have good reduction $v_p(4A^3+27B^2)=6k$
where $k\in\mathbb{Z}$.
Then the quadratic twist $y^2=x^3+p^{-2k}Ax+p^{-3k}B$
will work as long as $v_p(A)\ge 2k$ and $v_p(B)\ge 3k$.
Otherwise any quadratic twist making the discriminant a $p$-unit
will have coefficients which are non $p$-integral so no
quadratic twist will have good reduction.
The cases $p=3$ or $p=2$ will be harder :-)
ADDED Even in these awkward characteristics the
same argument shows that $v_p(4A^3+27B^2)$ being
a multiple of $6$ is a necessary condition.
Best Answer
To make things concrete, suppose that we're working in characteristic $\ne 2,3$ (we can do something similar in those cases, though it gets messier), and the equation of the curve is
$E : y^2 = x^3 + a x + b$
and the equation of the twist is:
$E^{(d)} : d y^2 = x^3 + a x + b$.
Multiplying this by $d^3$:
$(d^2 y)^2 = (d x)^3 + a d^2 (d x) + d^3 b$
for this to be isomorphic to $E$ over $k$ it is necessary and sufficient that there is a $\lambda \in k^*$ such that (see Silverman, or any other book on elliptic curves) such that $a d^2 = a \lambda^4$, $b d^3 = b \lambda^6$. If $a,b \ne 0$ then we must have $\lambda^6 = d^3$ and $\lambda^4 = d^2$, which shows that $\lambda^2 = d$ which you've ruled out. If $b=0$ you have, by taking square roots, $\lambda^2 = \pm d$, and only the minus sign is possible. If $a=0$ (and thus $b \ne 0$) then $\lambda^6 = d^3$. But then $(\lambda^3/d)^2 = d^3/d^2 = d$, which you've also ruled out. Thus, only $b=0$ is possible, which is $j=1728$.