Im in need to solve a large set of Exponential Diophantine Equations of the type $$\frac{1}{3} \left(\frac{1}{3} \left(-1+2^{2 (3 n-1)}\right) 2^{2 m-1}-1\right) 2^l\equiv 0\pmod3.$$
By inspection I arrive at the following solutions:
\begin{equation} \frac{1}{3} \left(\frac{1}{3} \left(-1+2^{2 (3 (3 n-2)-1)}\right) 2^{2 (3 m-1)-1}-1\right) 2^{2 l} \end{equation}
\begin{equation} \frac{1}{3} \left(\frac{1}{3} \left(-1+2^{2 (3 (3 n-1)-1)}\right) 2^{2 (3 m)-1}-1\right) 2^{2 l} \end{equation}
\begin{equation} \frac{1}{3} \left(\frac{1}{3} \left(-1+2^{2 (3 (3 n)-1)}\right) 2^{2 (3 m-2)-1}-1\right) 2^{2 l} \end{equation}
\begin{equation} \frac{1}{3} \left(\frac{1}{3} \left(-1+2^{2 (3 (3 n-2)-1)}\right) 2^{2 (3 m)-1}-1\right) 2^{2 l-1} \end{equation}
\begin{equation} \frac{1}{3} \left(\frac{1}{3} \left(-1+2^{2 (3 (3 n-1)-1)}\right) 2^{2 (3 m-2)-1}-1\right) 2^{2 l-1} \end{equation}
\begin{equation} \frac{1}{3} \left(\frac{1}{3} \left(-1+2^{2 (3 (3 n)-1)}\right) 2^{2 (3 m-1)-1}-1\right) 2^{2 l-1} \end{equation}
As the equations get longer, solving by inspection become too cumbersome, can anyone point me in a direction to systematically solve this kind of equation?
Tank you