10. Find the positive integer solutions of the equation $3^{x}=2^{x} y+1$.

Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly.
10. Solution: Rewrite the equation as $3^{x}-1=2^{x} y$
(1)
This indicates that if $(x, y)$ is a solution, then $x$ cannot exceed the exponent of the factor 2 in the standard factorization of $3^{x}-1$. Let $x=2^{m}(2 n+1)$, where $m, n$ are non-negative integers. Thus, we have
$$3^{x}-1=3^{2^{m}(2 n+1)}-1=\left(3^{2 n+1}\right)^{2^{m}}-1=\left(3^{2 n+1}-1\right) \prod_{k=0}^{m-1}\left(\left(3^{2 n+1}\right)^{2^{k}}+1\right)$$

Since $3^{2 n+1}=(1+2)^{2 n+1}$
$\equiv 1+2(2 n+1)+4 n(2 n+1)(\bmod 8)($ Binomial Theorem)
$\equiv 3(\bmod 8)$,
for the required exponent, it is 1 when $m=0$, and $m+2$ when $m=1,2, \cdots$. Therefore, we can conclude that $x$ cannot exceed $m+2$.

Thus, $2^{m} \leqslant 2^{m}(2 n+1)=x \leqslant m+2$. Hence, $m \in\{0,1,2\}$ and $n=0$.
From this, we can conclude that the positive integer solutions to the given equation are $(x, y)=(1,1),(2,2),(4,5)$.