I tried to discuss two cases when $m$ is even and odd. By discussing both cases, $m$ must be odd, and by using the fact that every odd perfect square is congruent to $1$ modulo $8$.
It is clear that $n=3$ and $m=3$ is a solution, and it seems that it is the only one? I tried to prove that the equation admits one solution, but I failed.
Can I get some help, and thanks in advance.
On
We have $2^n=(m+1)(m-1)$, so, by unique factorization, $m+1$ and $m-1$ must be powers of $2$.
But the only powers of $2$ that differ by $2$ are $4$ and $2$, so $m$ must be $3$.
On
Assume we are looking for solutions of the form 2^n=m^2-1=(m-1)(m+1). So this means that both m-1 and m+1 are divisible by 2. By inspection the integer solution is n=3 and m=-+3 since we have 2^3=8=(2)(4)=(-4)(-2). The other solutions are complex. Taking natural logs of both sides we have nln(2)=ln(|m^2-1|) so that n=ln(m^2-1)/ln(2) and ln(2) is irrational. Since if ln(2)=a then e^a=2, but e^a is transcendental. Here: Is the natural log of n rational? So only possible cases for integers solutions is when the numerator can be made of the form nln(2), so the ln(2) cancels this is indeed the case for m=-+3. https://www.wolframalpha.com/input/?i=2%5Ey%3D%28x-1%29%28x%2B1%29
Hint:
$$m^2-1=2^n$$
Check for $n=0,1,2$
As $m+1,m-1$ have same parity, both must be even for $n>2$
$$\dfrac{m+1}2\cdot\dfrac{m-1}2=2^{n-2}$$
Now as $\dfrac{m+1}2-\dfrac{m-1}2=1$
$$\left(\dfrac{m+1}2,\dfrac{m-1}2\right)=1$$
So, they must have opposite parities.
Now what is the odd factor of $2^{n-2}$