11. (2005 Balkan Mathematical Olympiad) Find the positive integer solutions of the equation $3^{x}=2^{x} y+1$.

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11. Rewrite the equation as
$$
3^{x}-1=2^{x} y \text {. }
$$

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 $3^{2 n+1}=(1+2)^{2 n+1} \equiv[1+2(2 n+1)+4 n(2 n+1)](\bmod 8) \equiv 3(\bmod 8)$, then
$$
\left(3 ^ { 2 n + 1 } 2 ^ { 2 ^ { k } } \equiv \left\{\begin{array}{l}
3(\bmod 8), \text { when } k=0 \text {; } \\
1(\bmod 8), \text { when } k=1,2, \cdots \text {; }
\end{array}\right.\right.
$$

Therefore, for the required exponent, it is 1 when $m=0$, and $m+2$ when $m=1,2, \cdots$. This allows us to conclude that $x$ cannot exceed $m+2$.
[From the above analysis, the right-hand side of equation (2) can be expressed as
$$
\begin{array}{l}
\quad(8 t+2)(8 t+4)\left(8 r_{1}+2\right)\left(8 r_{2}+2\right) \cdot \cdots \cdot\left(8 r_{m-1}+2\right) \\
=2^{m+2}(4 t+1)(2 t+1)\left(4 r_{1}+1\right)\left(4 r_{2}+1\right) \cdots \cdots \cdot\left(4 r_{m-1}+1\right) \\
=2^{x} y \text { (see equation (D) -- translator's note] }
\end{array}
$$

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