Consider the case of an elliptic curve $A$ over $\mathbb{Q}$. Let $A_{\ast}$ be the Weil restriction of $A_{K}$. Then $c(A_{\ast})=N_{K/\mathbb{Q}}(c(A_{K}))d_{K/\mathbb{Q}}^{2}$. Here $c$ is the conductor and $d$ the discriminant. As $A_{\ast}$ is isogenous to the product of $A$ and its twist, and isogenies don't change conductors, I think this gives what your want. Reference: Milne, Invent. Math. 1972, Pptn.1.

Consider the case of an elliptic curve $A$ over $\mathbb{Q}$ and a quadratic extension $K$ of $\mathbb{Q}$. Let $A_{\ast}$ be the Weil restriction of $A_{K}$. Then $c(A_{\ast})=N_{K/\mathbb{Q}}(c(A_{K}))d_{K/\mathbb{Q}}^{2}$. Here $c$ is the conductor and $d$ the discriminant. As $A_{\ast}$ is isogenous to the product of $A$ and its twist, and isogenies don't change conductors, I think this gives what your want. Reference: Milne, Invent. Math. 1972, Pptn.1.

The formula, in fact, holds for abelian varieties, so if a curve and its quadratic twist both have good reduction, then the quadratic field $K$ must be unramified.