$\newcommand{\Sha}{\operatorname{Sha}}$Let $E/\mathbb{Q}$ be an elliptic curve, and let $\Sha(E/\mathbb{Q})$ denote the Tate–Shafarevich group of $E/\mathbb{Q}$. It is known that the 2-primary component $\Sha(E/\mathbb{Q})[2]$ can become arbitrarily large as we vary $E$.

For example, Kramer found an elliptic curve over $\mathbb{Q}$ with discriminant $m(16m+1)$ and $\operatorname{\#Sha}(E/\mathbb{Q})[2] \geq 2^{2k-2}$, where $k$ is the number of prime factors of $16m+1$ ("A family of semistable elliptic curves with large Tate–Shafarevich groups," Proceedings of the AMS, 89 (1983)).

This elliptic curve has a discriminant $\Delta_E$ with a very large number of prime factors.

However, is it possible to enlarge the 2-part of the Tate–Shafarevich group while keeping the number of prime factors of the discriminant small? Specifically, can we increase $\dim_2\Sha(E/\mathbb{Q})[2] - \omega(\Delta_E)$ arbitrarily?

The largest value of $\dim_2\Sha(E/\mathbb{Q})[2] - \omega(\Delta_E)$ that I am aware of is 12. If anyone knows of an example with a larger $\dim_2\Sha(E/\mathbb{Q})[2] - \omega(\Delta_E)$, I would greatly appreciate the information.

Let $\omega(\Delta_E)$ denote the number of distinct prime factors of $\Delta_E$. The following two elliptic curves have a small number of prime factors of their discriminants, $\omega(\Delta_E) = 2$, and $\dim_2\Sha(E/\mathbb{Q})[2] - \omega(\Delta_E) = 4$:

1. $E_d: y^2 = x^3 + dx$ when we let $d$ be $p = 2^{2^k} + 1$, then $\Sha(E_{65537}) \cong (\mathbb{Z}/8\mathbb{Z})^2$. For more information, refer to the discussion on MathOverflow here.

2. Let $E_{p,n}: y^2 = x^3 + p^nx$ be an elliptic curve. According to LMFDB, for $(p,n) = (73,3)$, $\operatorname{\#Sha}(E_{p,n}) = 64$. This was mentioned in a MathOverflow discussion available here.

In Table 2 of the document available at Sza.pdf, the case of $(n,p) = (16,48)$ is cited as an example where $\Sha(E/\mathbb{Q}) - \omega(\Delta_E) = 18 - 6 = 12$.

This record of $\dim_2\Sha(E/\mathbb{Q})[2] - \omega(\Delta_E) = 12$ is the maximal one I have ever seen.