Find all positive integer solutions $(x, y, n)$ for the equation $1+2^{x}+2^{2 x+1}=y^{n}$.

Translate the text above into English, please keep the original text's line breaks and format, and output the translation result directly.
For a positive integer $m$, let $v_{2}(m)$ be the exponent of 2 in the prime factorization of $m$, then it is easy to know that $v_{2}\left(m_{1} m_{2}\right)=v_{2}\left(m_{1}\right)+v_{2}\left(m_{2}\right), m_{1}, m_{2} \in \mathbb{N}^{*}$
Since the left side of the equation is odd, $y$ must be odd.
(1) If $n$ is odd, then $x=v_{2}\left(y^{n}-1\right)=v_{2}(y-1)+v_{2}\left(y^{n-1}+y^{n-2}+\cdots+1\right)=v_{2}(y-$ so $y \geq 2^{x}+1$, thus $y^{3}>1+2^{x}+2^{2 x+1}=y^{n}$, then $n<3$
Therefore, $n=1, y=1+2^{x}+2^{2 x+1}$, i.e., $(x, y, n)=\left(t, 1+2^{t}+2^{2 t+1}, 1\right)$, where $t \in \mathbb{N}^{*}-$ points
(2) If $n$ is even, let $n=2 k, k \in \mathbb{N}^{*}$, it is verified that $x \neq 1$ $y^{2 k}=1+2^{x}+2^{2 x+1}=\left(1+2^{x-1}\right)^{2}+7 \cdot 2^{2 x-2}$, i.e., $\left(y^{k}+1+2^{x-1}\right)\left(y^{k}-1-2^{x-1}\right)=7 \cdot 2^{2 n-2}$ Since $x \geq 2,\left(y^{k}+1+2^{x-1}\right)-\left(y^{k}-1-2^{x-1}\right)=2\left(1+2^{x-1}\right) \equiv 2(\bmod 4)$

So there are only the following four cases: points
(1) $\left\{\begin{array}{l}y^{k}+1+2^{x-1}=14 \\ y^{k}-1-2^{x-1}=2^{2 x-3}\end{array}\right.$, then $1+2^{x-1}=7-2^{2 x-4}$, no positive integer solution
(2) $\left\{\begin{array}{l}y^{k}+1+2^{x-1}=2^{2 x-3} \\ y^{k}-1-2^{x-1}=14\end{array}\right.$, then $1+2^{x-1}=2^{2 x-4}-7$, solving gives $x=4$

Thus, $(x, y, n)=(4,23,2)$
(3) $\left\{\begin{array}{l}y^{k}+1+2^{x-1}=7 \cdot 2^{2 x-3} \\ y^{k}-1-2^{x-1}=2\end{array}\right.$, then $1+2^{x-1}=7 \cdot 2^{2 x-4}-1$, no positive integer solution
(4) $\left\{\begin{array}{l}y^{k}+1+2^{x-1}=2 \\ y^{k}-1-2^{x-1}=7 \cdot 2^{2 x-3}\end{array}\right.$, then $1+2^{x-1}=1-7 \cdot 2^{2 x-4}$, no positive integer solution

In summary, $(x, y, n)=(4,23,2)$ or $\left(t, 1+2^{t}+2^{2 t+1}, 1\right)$, where $t \in \mathbb{N}^{*}$