Is there an elliptic curve $E/\mathbb{Q}$ such that for every quadratic number field $K$, the rank of $E$ over $K$ is equal to the rank of $E$ over $\mathbb{Q}$?
Consider $K = \mathbb{Q}(\sqrt{D})$. The rank of $E$ over $K$ is equal to the sum of the ranks of $E$ over $\mathbb{Q}$ and $E_D$ over $\mathbb{Q}$.
Assuming the Birch and Swinnerton-Dyer (BSD) conjecture for elliptic curves over $\mathbb{Q}$ holds true, results by Mazur and Rubin suggest the existence of an elliptic curve for which $\text{rank}(E/\mathbb{Q}(\sqrt{D})) = \text{rank}(E/\mathbb{Q}) = 1$.
I would appreciate any insights on this topic, especially those not reliant on the BSD conjecture.
Assuming finiteness of Tate-Shafarevich groups, there is no such elliptic curve. Indeed, you can always find a $D$ such that the root number of $E_D$ is $-1$; and then the $p$-parity conjecture for any prime $p$ (which is known for elliptic curves over $\mathbb{Q}$, thanks to work of many people; I will let you google it) together with finiteness of the $p$-part of the Tate-Shafarevich group of $E_D$ (which is not known, but widely conjectured) implies that the rank of $E_D$ is odd.
We expect half of all quadratic twists to have rank $0$, half to have rank $1$, and $0$% to have higher rank. This is the Goldfeld conjecture. Its analogue for $2^\infty$-Selmer ranks is known in most cases, thanks to the groundbreaking work of Alexander Smith. If $2$-parts of Tate-Shafarevich groups are finite, then ranks are the same as $2^\infty$-Selmer ranks.
EDIT:
If $E$ is an elliptic curve over a number field $K$ and $m\geq 2$ is an integer, then the set of $L/K$ with $[L:K] = m$ and $\textrm{rank}E(L) > \textrm{rank}E(K)$ is infinite.
This is a consequence of Siegel's theorem and Hilbert's irreducibility theorem; see https://arxiv.org/abs/1209.0933
For $m=2$ (and also $m=3$) this statement is simple to prove; see the proof of Corollary 2 in loc. cit.