Find the positive integers $m,n$ such that $2^{m}\cdot3^{n}-1$ is a perfect square. Find the positive integers $m,n$ such that $2^{m}\cdot3^{n}-1$ is a perfect square.
I tried to let $2^{m}\cdot3^{n}-1=x^2$ then $x$ is an odd number but here i don't have any ideas. I think it can use Fermat-Euler theorem. Is it a right way? How can i solve it?
 A: Since $m,n$ are positive then the number $2^m3^n-1$ will be of the form $6k-1 $ or $6k+5$. Now notice that no square is of the form $6k+5$.
A:   First claim: $n \leq 0$.
Suppose on contraary that $1 \leq n$, so then we have:
$2^m3^n-1\overset{3}{\equiv}0-1\overset{3}{\equiv}-1$,
which contradicts with the fact that, the only possible values for squares module $3$, are $0$ and $1$.
  Socend claim: $m \leq 1$
Suppose on contraary that $2 \leq m$, so then we have:
$2^m3^n-1\overset{4}{\equiv}0-1\overset{4}{\equiv}-1$,
which contradicts with the fact that, the only possible values for squares module $4$, are $0$ and $1$.


So the only possibilities for $2^m3^n-1$ to be square, are the below cases:
$n=0, m=0$;
$n=0, m=1$.
The first case gives us $x=0$, and the second gives $x=1$.
A: Let's assume
$$2^m3^n-1=x^2 \tag{1}$$
If we assume $n>0$ then $\gcd(x, 3)=1$ (otherwise $3 | 1$), thus, from Euler's theorem
$$x^2 \equiv 1 \pmod{3}$$
but from $(1)$:
$$x^2 \equiv -1 \pmod{3} \Rightarrow 0\equiv 2 \pmod{3}$$
which is a contradiction, so $n=0$.
Now we look at
$$2^m-1=x^2 \tag{2}$$
and assume $m\geq1$. As you already mentioned, $x$ must be odd, e.g. $x=2q+1$, then
$$2^m-1=(2q+1)^2 \Rightarrow 2^{m-1}=2q(q+1)+1$$
which limits $m$ to $1$.
Altogether $m \in \{0,1\}, n=0$.
