An element $\alpha \in H^i(k;\mu_2)$ is called a pure symbol in case it has the form $\alpha = (a_1) \cup\dots\cup (a_i)$, $(a_i) \in H^1(k;\mu_2)$.
It is known that for every pure symbol $\alpha $ there is a smooth, projective variety $X_\alpha$ such that over the function field $k(X_\alpha)$, the pure symbol $\alpha$ becomes zero.
Assume we are given two pure symbols $\alpha, \beta \in H^i(k;\mu_2)$. Then the following is known to hold:
If $\alpha$ becomes trivial over $k(X_\beta)$ and $\beta$ becomes trivial over $k(X_\alpha)$, then $\alpha$ and $\beta$ are isomorphic.
In case $i = 1$ I'd say that it is enough to conclude that $\alpha$ and $\beta$ are isomorphic, in case we know that $\alpha$ becomes trivial over $k(X_\beta)$.
Question: Is that true for any $i$ ?
One could restate this as: Can an $i$-Pfister form become trivial over the generic point of another $i$-Pfister form, without the two being Witt equivalent ?
It is shown in corollary 23.6 of The Algebraic and Geometric Theory of Quadratic Forms (by Elman, Karpenko and Merkurjev) that if $\varphi$ is a non-trivial $i$-Pfister form and $\psi$ is any anisotropic form of even dimension, then if $\psi$ becomes hyperbolic over the function field of $\phi$, there is an isometry $\psi\simeq b\cdot \varphi$ for some bilinear form $b$.
If in particular $\psi$ is itself an $i$-Pfister form, then we must have $b=1$, so $\psi\simeq \varphi$.