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.

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 $y = 0.$
