4. (2003 Turkey Mathematical Olympiad) Find the triple positive integer solutions $(x, m, n)$ of the equation $x^{m}=2^{2 n+1}+2^{n}+1$.
4. Solution: Clearly, $x$ is an odd number, and let the power of 2 in $t$ be $V_{2}(t)$.

If $m$ is odd, let $y=x-1$, then $x^{m}-1=(y+1)^{m}-1=y^{m}+C_{m}^{1} y^{m-1}+C_{m}^{2} y^{m-2}+\cdots+C_{m}^{m-1} y$.

In the term $C_{m}^{m-1} y$, the power of 2 is the power of 2 in $y$, and the other terms satisfy $V_{2}\left(C_{m}^{i} y^{i}\right)=V_{2}(y)+(i-1) V_{2}(y)+V_{2}\left(C_{m}^{i}\right)>V_{2}(y)$, so $V_{2}\left(x^{m}-1\right)=V_{2}(y)=V_{2}(x-1)$.

Also, $V_{2}\left(x^{m}-1\right)=V_{2}\left(2^{2 n+1}+2^{n}\right)=n$, thus $V_{2}(x-1)=n$.
Then $2^{n} \mid x-1$. Therefore, $x-1 \geqslant 2^{n}, x \geqslant 2^{n}+1$.
And $x^{3} \geqslant\left(2^{n}+1\right)^{3}=2^{3 n}+3 \times 2^{2 n}+3 \times 2^{n}+1>2^{2 n+1}+2^{n}+1=x^{m}$.
So, $m<3$. Hence $m=1$, at this time $x=2^{2 n+1}+2^{n}+1$.
If $m$ is even, let $m=2 m_{0}$. Then $\left(x^{m_{0}}\right)^{2}=7 \times 2^{2 n-2}+\left(2^{n-1}+1\right)^{2}$,
which means $\left(x^{m_{0}}-2^{n-1}-1\right)\left(x^{m_{0}}+2^{n-1}+1\right)=7 \times 2^{2 n-2}$.
If $n=1$, then $x^{m}=2^{3}+2^{1}+1=8+2+1=11$ is not a square number, which is impossible.
Therefore, $n \geqslant 2$.
So, $x^{m_{0}}-2^{n-1}-1 \not\equiv x^{m_{0}}+2^{n-1}+1(\bmod 4)$.
Since they are both even, one of them has a power of 2 of 1, and there are only the following four cases:
$$\begin{array}{l}
\left\{\begin{array}{l}
x^{m_{0}}-2^{n-1}-1=14 \\
x^{m_{0}}+2^{n-1}+1=2^{2 n-3}
\end{array}\right. \\
\left\{\begin{array}{l}
x^{m_{0}}-2^{n-1}-1=2^{2 n-3}, \\
x^{m_{0}}+2^{n-1}+1=14 ;
\end{array}\right. \\
\left\{\begin{array}{l}
x^{m_{0}}-2^{n-1}-1=7 \times 2^{2 n-3} \\
x^{m_{0}}+2^{n-1}+1=2 ;
\end{array}\right. \\
\left\{\begin{array}{l}
x^{m_{0}}-2^{n-1}-1=2, \\
x^{m_{0}}+2^{n-1}+1=7 \times 2^{2 n-3}
\end{array}\right.
\end{array}$$

From (1), we have $2^{n-1}+1=2^{2 n-4}-7$, i.e.,
$2^{n-4}+1=2^{2 n-7}$, solving gives $n=4, x^{m_{0}}=23$. Hence $x=23, m_{0}=1$.
From (2), we have $2^{n-1}+1=7-2^{2 n-4}$, i.e., $2^{n-1}+2^{2 n-4}=6$.
Since $6=4+2$, there is no solution.
From (3), we have $2^{n-1}+1=1-7 \times 2^{2 n-4}$, which is also impossible.
From (4), we have $2^{n-1}+1=7 \times 2^{2 n-4}-1$, i.e., $2^{n-2}+1=7 \times 2^{2 n-5}$.
Considering the number of 1s in the binary representation, it is also impossible.
Therefore, $(x, m, n)=\left(2^{2 n+1}+2^{n}+1,1, n\right),(23,2,4)$.