Bayesian Nash Equilibrium in a Duopoly Cournot Competition
If the capacities $q_1$ and $q_2$ are the same, say $\overline{q}$ you can rule out some cases.
The best response functions are given by:
$$
\begin{align*}
\hat q_1 &amp;= \min\left\{\frac{M - (1-\theta) \hat q_2^L - \theta \hat q_2^H - m}{2}, \overline{q}\right\},\\
\hat q_2^L &amp;= \min\left\{\frac{M - \hat q_1 - m}{2}, \overline{q}\right\},\\
\hat q_2^H &amp;= \min\left\{\frac{\alpha M - \hat q_1 - m}{2}, \overline{q}\right\}
\end{align*}
$$
First we can show that $\hat q_2^H \ge \hat q_2^L$.
Proposition: $\hat q_2^H \ge \hat q_2^L$

Proof: If $\hat q_2^H = \overline{q}$ the proof is obvious. Else, from $\alpha &gt; 0$, we have:
$$
\hat q_2^H = \frac{\alpha M - \hat q_1 - m}{2} \ge \frac{M - \hat q_1 - m}{2} \ge q_2^L.
$$

Next, we can also show that $\hat q_2^L \ge \hat q_1$.
Proposition: $\hat q_2^L \ge \hat q_1$.

Proof: if $\hat q_2^L = \overline{q}$ the proof is obvious. If not, then:
$$
\begin{align*}
\hat q_1 - \hat q_2^L &amp;\le \frac{M - (1-\theta) \hat q_2^L - \theta \hat q_2^H - m}{2} - \hat q_2^L,\\
&amp;= \frac{M - (1-\theta) \hat q_2^L - \theta \hat q_2^H - m}{2} - \frac{M - \hat q_1 - m}{2},\\
&amp;= \frac{-(1-\theta) \hat q_2^L - \theta \hat q_2^H + \hat q_1}{2},\\
\to 2 \hat q_1 - 2 \hat q_2^L &amp;\le -(1- \theta) \hat q_2^L - \theta \hat q_2^H + \hat q_1 \\
\to \hat q_1 &amp;\le \hat q_2^L - \theta(\hat q_2^H - \hat q_2^L) \le \hat q_2^L.
\end{align*}
$$
where the last line follows from $\hat q_2^H \ge \hat q_2^L$.

The two propositions show that we only have to consider the following cases:

$\hat q_2^H &lt; \overline{q}$ and $\hat q_2^L &lt; \overline{q}$ and $\hat q_1 &lt; \overline{q}$
$\hat q_2^H = \overline{q}$ and $\hat q_2^L &lt; \overline{q}$ and $\hat q_1 &lt; \overline{q}$.
$\hat q_2^H = \hat q_2^L = \overline{q}$ and $\hat q_1 &lt; \overline{q}$.
$\hat q_2^H = \hat q_2^L = \hat q_1 = \overline{q}$.

For each of them you can compute the equilibrium values and then you need to check if they indeed correspond to the best responses.
This reduces the number of cases from 8 to 4. I don't think you can do better.