Proving that there are no other integer solutions The integer solutions for this equation:
$m ^ 2 = 2 \cdot 3 ^ n - 5$ 
are these four couples: $(n; m): (1;-1) (1;1) (3;7) (3;-7)$. 

How can I prove there are no other solutions?


Here is an approach.
Since $m$ is odd, let $m = 2k + 1$
We have $4k^2 + 4k + 6 = 2*3^n \implies 2k^2 + 2k + 3 = 3^n \implies 2k(k+1) = 3(3^{n-1} - 1)$
So $3$ has to divide $k$ or $k+1$ and $3$ cannot be a factor of both $k$ and $k+1$. 
If $3$ divides $k$, then $k$ divides $3 \implies k = 3 \implies m = 7, n = 3$
If $3$ divides $k$ and $k+1$ divides $3 \implies k = 0, m = 1, n = 1$
