It is not possible to prove that any surface $S\subset\mathbb{R}^3$ that has both $H$ and $K$ nonconstant but $H^2-K=c^2$ for some constant $c>0$ must have $K$ be negative. In fact, one cannot conclude anything about the sign of $K$ from only these hypotheses, beyond the fact that $K\ge -c^2$.

In fact, it's not difficult to show, using the Cartan-Kähler Theorem, that there exist plenty of such surfaces with $K$ positive. I can supply the details if you want.

Whether adding the condition that the surface be a Bonnet surface could force $K$ to be negative is another question. I think that is probably not the case, but I'd have to work it out to be sure.

**Added after the question was changed:** There are *no* Bonnet surfaces $S\subset\mathbb{R}^3$ with the property that $H$ is nonconstant while $H^2-K>0$ is constant (where 'Bonnet surface' means that there is a nontrivial isometric deformation of $S$ in $\mathbb{R}^3$ that preserves the mean curvature $H$). Thus, the OP's question is not sensible.

A proof that such surfaces do not exist can be got by simply differentiating the equation $H^2-K = r^2$ (where $r>0$ is a constant) several times and combining it with the three equations that follow from requiring that the surface admit a nontrvial $1$-parameter family of isometric embeddings into $\mathbb{R}^3$, all with the same *nonconstant* mean curvature function $H$.

A careful analysis shows that this overdetermined system is incompatible, even locally. The full analysis is too long to input here. However, you can find a discussion of this in Section 7.2 of my introductory lectures on exterior differential systems. If you follow those notes to the discussion on page 50, what this amounts to is seeking a solution to the ODE system on $\mathbb{R}^5$ near the bottom of that page that also satisfies $r_1=0$ while $H_1\not=0$ and $H$ and $r$ are nonzero. Since there are no such solutions, there are no such surfaces.