Solving the Diophantine Equation $2 \cdot 5^n = 3^{2m} + 1$ over $\mathbb{Z}^+$. 
Prove that $(m, n) = (1, 1)$ is the only solution for the Diophantine Equation $$2 \cdot 5^n = 3^{2m} + 1$$ where $(m, n) \in (\mathbb{Z}^+)^2$.

I've managed to prove that both $m$ and $n$ are odd seeing $\bmod 3\text{ and } 10$ respectively. Also, $\forall n \ge 1$, $10$ divides the LHS. I am not able to proceed from here. Any help would be appreciated.

Here is an elementary solution. $n = 1$ gives the solution above. Now, suppose $ n > 1$. Thus, $9^m + 1 \equiv 0 \pmod {25}$. Looking at all the possible cases, we have $m = 10k + 5$ for integer $k$. We can check that there are no solutions for $m = 2, \ldots, 5$. Thus, $k > 0$. Theefore,
$$ 2 \cdot 5^n = 3^{2(10k + 5)} + 1 = 3^{10} \cdot 3^{20k} + 1.$$
Now $1181$ divides $3^{20} - 1$. Thus, $3^{20} \equiv 1 \pmod {1181}$ and $3^{10} \equiv - 1 \pmod{1181}$. Now taking the equation $\pmod{1181}$ gives 
$$2 \cdot 5^n \equiv -1 \cdot 1 + 1 \equiv 0 \pmod {1181}$$ which has no solutions. Thus $(m,n) = (1,1)$ is the only solution.
