In Number Theory by Boreivich, he states in the Algebraic Supplement:
1. Quadratic Forms over Arbitrary Fields of Characteristic ≠ 2
... Throughout, K will denote an arbitrary field whose characteristic is not 2. [page 390]
and later
This equation, which can be written in the form
ξ2 - (γξ + 1)2 = 0,
always has a solution ξ0 ∈ K for any γ ∈ K (recall that the characteristic of K is not 2). [Page 393]
It's unclear to me how a characteristic of any value other than 2 guarantees a solution to the quadratic in ξ.
In case the question is unclear, here is context:
In any field, the equation $\xi^2-(\gamma\xi+1)^2=0$ is equivalent to the disjunction of $\xi=\gamma\xi+1$ and $\xi=-\gamma\xi-1$, or equivalently, $\xi=\frac{1}{1-\gamma}$ and $\xi=\frac{-1}{1+\gamma}$. Thus, it has a solution iff $1-\gamma\neq 0$ or $1+\gamma\neq 0$.
This is always true, unless $\gamma=1$ in a field of characteristic $2$.
Putting it differently, the equation is equivalent to $$ \xi^2(1-\gamma^2)+\xi(-2\gamma)-1=0. $$ If $1-\gamma^2=0$, and the characteristic is not $2$, then this is a non-degenerate linear equation, so it has a unique solution.
If the characteristic is not 2 and $1-\gamma^2\neq 0$, this is a non-degenerate quadratic equation with discriminant $4\gamma^2+4(1-\gamma^2)=4$, which always has a square root, so the equation has two distinct solutions.
In characteristic 2, the equation reduces to $$ \xi^2(1-\gamma^2)-1=0, $$ which is easily checked to be equivalent to $$ \xi(\gamma+1)=1, $$ so it has a (unique) solution iff $\gamma\neq 1$.