I think this question presupposes something that is incorrect. Here is an example. Fix a field $F$ of characteristic 2 and put $Z$ for the field $F(a,b)$ where $a$ and $b$ are indeterminates. Define $D$ to be the quaternion division algebra with center $Z$ generated by elements $i, j$ satisfying
$$
i^2 = a,\quad j^2 + j = b,\quad ij = (j+1)i.
$$
Let us denote it by $\lt a,b]$, as PK Draxl does in his book "Skew Fields". This is a *division* algebra because $a$ is not a norm from the separable quadratic extension $E$ obtained from $Z$ by adjoining a root of $x^2 + x + b$.

Obviously $D$ contains the purely inseparable quadratic extensions $Z(\sqrt{a})$. I claim it also contains the extension $Z(\sqrt{ab})$. To see this, we calculate in the Brauer group:
$$
\lt a,b] = \lt ab^2,b] = \lt ab,b] + \lt b,b] = \lt ab, b]
$$
where the last equality is because $\lt b,b]$ is split, i.e. isomorphic to 2-by-2 matrices, which follows from the fact that $b$ itself is a norm from $E$ (in fact, the norm of the element $x$).

Because $b$ is a nonsquare in $K$, we have found two non-isomorphic purely inseparable quadratic extensions in $D$ of exponent 1.