$H$ is a hermitian $4\times 4$ matrix, and after squaring twice I arrived at the following equation to solve:
$$(H^2-\alpha\Bbb1)^2=4\beta\Bbb1, \quad\text{where }\Bbb1 \text{ is the identity matrix and }\alpha,\beta \geq0.$$
According to the paper I'm reading the solutions for the eigenvalues of $H$ are:
$$E^2_\pm=\alpha\,\pm 2\sqrt{\beta}$$
At least to get a similar form you could take a square root of the equation to get $$H^2=\left(\alpha\pm2\sqrt{\beta}\right)\cdot\Bbb1$$
Now I think this might be justified as follows: $H$ is hermitian so it's eigenvalues are real, hence the square of these must be positive or zero. $\alpha+2\sqrt{\beta}\geq0$ always, and we would only allow $\alpha-2\sqrt{\beta}$ if it is positive also. Although this doesn't really explain why other square roots of the identity matrix aren't allowed, like $-\Bbb1$.
The only extra information I believe is relevant is that $H$ isn't proportional to the identity matrix, but couldn't there in principle be repeated eigenvalues?
The only eigenvalue of $(H^2 - \alpha I)^2$ is obviously $4\beta$. Now, if $A$ is a square matrix then the eigenvalues of $A^2$ are given by the squares of all eigenvalues of $A$. Therefore (going back), the only possible eigenvalues of $H^2 - \alpha I$ are $\pm 2\sqrt\beta$. But note that it could be only one of them. Now, again, if $A$ is a square matrix, then the eigenvalues of $A - \alpha I$ are given by $\lambda_i - \alpha$, where the $\lambda_i$ are the eigenvalues of $A$. Thus, the only possible eigenvalues of $H^2$ are $\alpha\pm 2\sqrt\beta$. These are two guys. If one of them is negative, it cannot be an eigenvalue of $H^2$. If both are positive, still only one could be an eigenvalue of $H^2$. Finally, the only possible eigenvalues of $H$ are $\pm\sqrt{\alpha\pm 2\sqrt\beta}$, depending on whether the square root exists.