Can we prove that the equation $2^n+1=5^m$ has no integral solution other than $(2, 1)$? $\textrm{Assume, on the contrary, that }n\geq 4,\text{then }$
\begin{aligned} 2^{n} &=5^{m}-1 \\ &=(5-1)\left(5^{m-1}+5^{m-2}+\cdots+1\right) \\ &=4\left(5^{m-1}+5^{m-2}+\cdots+1\right) \\ 2^{n-2} &=5^{m-1}+5^{m-2}+\cdots+1 \\ 2^{n-2} & \equiv \underbrace{1+1+\cdots+1}_{m \text { ‘1’s }}(\bmod 4) \\ 0 & \equiv m \quad(\bmod 4) \\ \therefore \quad m &=4 k \text { for some integer } k . \end{aligned}
$\text {Again, }$
$\begin{aligned}2^{n} &=5^{4 k}-1 \\ \displaystyle \quad &=\left(5^{4}\right)^{k}-1 \\ \displaystyle \quad &=\left(5^{4}-1\right)\left(625^{k-1}+625^{k-2}+\cdots+1\right) \\ \displaystyle \quad &=624\left(625^{k- 1}+625^{k-2}+\cdots+1\right) \\&\Rightarrow  624 | 2^{n},\text {which leads to a contradiction. } \\\therefore \quad n=0,1,2 \text{ or }3.\end{aligned}$
After checking one by one, we can conclude that $(n, m)=(2,1)  $
is the unique integer solution.
Your opinions and alternative proofs are highly appreciated.
 A: We suspect the largest solution is $4+1 = 5.$  Therefore write $2^n - 4 = 5^m - 5.$  Make new letters $x +2 = n$  and $y+1 = m. $   This becomes
$$ 4 (2^x - 1)  = 5 (5^y-1)  $$
ASSUME  that $x \geq 1,  y \geq 1.$
Since $2^x \equiv 1 \pmod 5,$  we see that $4 | x.$  Furthermore $2^4 - 1 |  2^x - 1.$  And $15 = 3 \cdot 5.$  So far, $3 | 2^x - 1.$  This means $3 | 5^y - 1$
Alright,   $5^y \equiv 1 \pmod 3.$  This requires $2 | y.$ $5^2 - 1 = 24 = 3 \cdot 8.  $ Thus $8 | 5^y - 1$  and $ 8 | 4(2^x - 1)$
The final $ 8 | 4(2^x - 1)$   is a contradiction: if $x \geq 1$   then $2^x - 1$  is odd.
Therefore $x=0$  and $y = 0.$
A: Another way to do this. Assume that $n \ge 3$. Then for the equation $5^m \pmod {2^n} = 1$ to hold for positive integral $m$, the equation $5^m \pmod 8 = 1$ must hold, which gives $m$ even, and $n \ge 3$ gives $m$ positive as well.
So it now remains to show that there are no positive integral [in $n$ and $m$] solutions to the equation
$$2^n = 5^m-1 = (5^k-1)(5^k+1),$$
where $n$ is an integer at least $3$, and $m$ is an even and positive integer, and thus $k = \frac{m}{2}$ is a positive integer i.e., difference of two squares.
If there were such solutions, both $5^k-1$ and $5^k+1$ would have to be powers of $2$, so let $b$ and $a$ be integers such that $5^k+1=2^b$ and $5^k-1=2^a$. Note however, that $(5^k+1)-(5^k-1)=2$, and that $2^b-2^a= 2^a(2^{b-a}-1)$, is $2$, only if $a=1$ and $b=2$. However, there is no integral $k$ that gives $5^k+1=4$ and $5^k-1=2$. So indeed, there are no solutions to the equation
$$2^n = (5^k-1)(5^k+1),$$
where $n$ is an integer at least $3$, and $k$ is a positive integer. Thus from the top paragraph, there are no nontrivial positive integral solutions in $m,n$ to $2^n+1=5^m$ with $n \ge 3$ that is.

The trick of establishing that [whenever the problem allows it that is], all solutions must satisfy an equation of the form $$(A-c)(A+c) = k^n$$ for some integral $k \ge 2$ and some small $c$, is used often. If something like this can be established, then one can conclude from this that the possible values that $n$ may take, falls into a [VERY] *bounded set, which leaves us only a few possible choices to go through. Above we established above, for example, that both $5^k-1$ and $5^k+1$, which differ by only $2$, must both be powers of $2$. That ended up leaving few choices for $k$ and $n$. Indeed, for two integers $A=5^k+1$ and $A-2=5^k-1$ that differ by only $2$, to both be powers of $2$, both $A$ and $A-2$ must be quite small, and thus in turn, so must $A=5^k+2$ and $A-2=5^k-2$, and thus in turn, so must their product $2^n$. Then as $n$ must be small, this reduces finding all the solutions to the Diophantine equation, to checking only the choices inside a very small set.
