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 $a=0$ then by taking cube roots (and thus $\lambda^2 = b \omega d$, where ne 0$) then $\omega$ is a primitive cube root of 1\lambda^6 = d^3$. However, the square root of a primitive cube root of 1 is another primitive cube root of 1 (in fact its square) so that would imply that But then $(\lambda^3/d)^2 = d^3/d^2 = d$is a square, which you've also ruled out. Thus, only $b=0$ is possible, which is $j=1728$.
