Consider real numbers $m,n,p,q\ge 0$ and the following two matrices
$$M_1= \left(\begin{matrix} 0&0&1\\0&m&n\\1&n&q \end{matrix} \right), \ \ M_2= \left(\begin{matrix} 0&1&0\\1&p&m\\0&m&n \end{matrix} \right).$$
Let $\chi_i$ be the characteristic polynomials of $M_i$, then:
$$\chi_1(x) = -(x^3-(q+m)x^2+(qm-1-n^2)x+m)$$
$$\chi_2(x) = -(x^3-(p+n)x^2+(pn-1-m^2)x+n)$$
Consider the case where these matrices commute, which is equivalent to $$(\star)\qquad m^2 + n^2 = mq +np + 1.$$
Remark: The relation $(\star)$ and the polynomials $\chi_i$ already appeared in this answer of Max Alekseyev.
Let us assume that $m \le n$. Then the above equation forces $n>0$.
Then we can take $p=(m^2 + n^2 - mq - 1)/n$.
Then, these matrices are self-adjoint and commuting, so simultaneously diagonalizable. Thus, there is an invertible matrix $P$ and (real) eigenvalues $\alpha_i, \beta_i, \gamma_i$ (the roots of $\chi_i$) such that:
$$P^{-1}M_iP= \left(\begin{matrix} \alpha_i&0&0\\0&\beta_i&0\\0&0&\gamma_i \end{matrix} \right)$$ By positivity $\max(\alpha_i, \beta_i, \gamma_i) = \Vert M_i \Vert$. Let us assume that $\Vert M_i \Vert = \gamma_i$.
We can also assume (up to permutation) that $\alpha_1 = min(\alpha_1, \beta_1, \alpha_2, \beta_2)$.
Question: For which $(m,n,q)$ we have $\frac{\alpha_1^3}{\gamma_1} + \frac{\alpha_2^3}{\gamma_2} + 1 <0$ ?
Investigation
I used SageMath to make some computation when $q=0$, and it seems that if $m < n$ then the expected property holds.
Just with $(m,n,q) = (1,n,0)$, SageMath provides complicated formula for $\alpha_i, \beta_i, \gamma_i$:
$\alpha_1 = -\frac{1}{3}(\frac{1}{2})^{2/3}(3n^2 + 4)\zeta u^{-1/3} - \frac{1}{6}(\frac{1}{2})^{1/3}u^{1/3} \overline{\zeta} + 1/3$
$\beta_1 = -\frac{1}{3}(\frac{1}{2})^{2/3}(3n^2 + 4) \overline{\zeta} u^{-1/3} - \frac{1}{6}(\frac{1}{2})^{1/3}u^{1/3}\zeta + 1/3$
$\gamma_1 = \frac{2}{3}(\frac{1}{2})^{2/3}(3n^2 + 4)u^{-1/3} + \frac{1}{3}(\frac{1}{2})^{1/3}u^{1/3} + 1/3$
$\alpha_2= -\frac{1}{3}(\frac{1}{2})^{2/3}(n^2 + 6) \overline{\zeta} v^{-1/3} - \frac{1}{6}(\frac{1}{2})^{1/3}v^{1/3}\zeta + \frac{2}{3}n$
$\beta_2= -\frac{1}{3}(\frac{1}{2})^{2/3}(n^2 + 6)\zeta v^{-1/3} - \frac{1}{6}(\frac{1}{2})^{1/3}v^{1/3} \overline{\zeta} + \frac{2}{3}n$
$\gamma_2= \frac{2}{3}(\frac{1}{2})^{2/3}(n^2 + 6)v^{-1/3} + \frac{2}{3}n + \frac{1}{3}(\frac{1}{2})^{1/3}v^{1/3}$
with $\zeta = (i\sqrt{3} + 1)$, $u = 9n^2 + 9in(\frac{4}{3}n^4 + \frac{13}{3}n^2 + \frac{32}{3})^{1/2} - 16$ and $v = 16n^3 - 18(n^2 - 2)n - 27n + 9i(\frac{4}{3}n^4 + \frac{13}{3}n^2 + \frac{32}{3})^{1/2}$.
We observe with a computer and large $n$ that $\frac{1}{n}(\frac{\alpha_1(n)^3}{\gamma_1(n)} + \frac{\alpha_2(n)^3}{\gamma_2(n)} + 1) \to -2$
Example: for $n=2$ we have $$\left(\begin{matrix} \alpha_1 & \beta_1 & \gamma_1\\ \alpha_2 & \beta_2 & \gamma_2 \end{matrix} \right) \simeq \left(\begin{matrix} -1.903211925911 & 0.1939365664746 & 2.709275359436 \\ 1.311107817465 & -0.4811943040920 & 3.170086486626 \end{matrix} \right)$$ and $\frac{\alpha_1(n)^3}{\gamma_1(n)} + \frac{\alpha_2(n)^3}{\gamma_2(n)} + 1 \simeq -0.83357599939<0.$
If we plot for $n=2,3,\dots, 50$, we get
Now it seems complicated to use such formulas (at least by hand) to answer the question.
Here is a necessary condition: $\chi_1(-m^{1/3}) <0$.
Indeed, since. $\alpha_2>0$, we have $$\frac{\alpha_1^3}{\gamma_1} + 1 < \frac{\alpha_1^3}{\gamma_1} + \frac{\alpha_2^3}{\gamma_2} + 1 <0,$$ implying that $\alpha_1^3<-\gamma_1<-m$. That is, $\chi_1(-m^{1/3}) <0$.
Just in case, here is a couple of inequalities equivalent to the one in question: $$\gamma_2(\alpha_1^3+m)+\gamma_1(\alpha_2^3+n)<0,$$ $$(\frac{\gamma_1}{\alpha_1^3}+1)(\frac{\gamma_2}{\alpha_2^3}+1) > 1.$$