Solution:
The equation is equivalent to
$$(3 x)^{2}+2 \cdot 3^{2 y+1}=3 x\left[2^{(2 y+1)+1}-1\right]$$
Letting $a=3 x$ and $b=2 y+1$, we have
$$a^{2}+2 \cdot 3^{b}=a\left(2^{b+1}-1\right)$$
Case 1: $b=1$
We have $a^{2}-3 a+6=0$ which has no integer solution for $a$. Thus, there is no solution in this case.

Case 2: $b=3$
We have $a^{2}-15 a+54=0$, whose roots are 6 and 9, both divisible by 3. This case then has the solutions $(2,1)$ and $(3,1)$ for the original equation.

Case 3: $b=5$
We have $a^{2}-63 a+486=0$, whose roots are 9 and 54, both divisible by 3. We get as additional solutions $(3,2)$ and $(18,2)$.

Case 4: $b \geq 7$, $b$ odd
It follows from (1) that $a \mid 2 \cdot 3^{b}$. Since $a$ is divisible by 3, either $a=3^{p}$ for some $1 \leq p \leq b$ or $a=2 \cdot 3^{q}$ for some $1 \leq q \leq b$.

For the first case $a=3^{p}$, let $q=b-p$. Then we have
$$2^{b+1}-1=a+\frac{2 \cdot 3^{b}}{a}=3^{p}+2 \cdot 3^{q}$$
For the second case $a=2 \cdot 3^{q}$, let $p=b-q$. Then we have
$$2^{b+1}-1=a+\frac{2 \cdot 3^{b}}{a}=2 \cdot 3^{q}+3^{p}$$
In either case,
$$2^{b+1}-1=3^{p}+2 \cdot 3^{q}$$
where $p+q=b$. Consequently, $2^{b+1}>3^{p}$ and $2^{b+1}>2 \cdot 3^{q}$. Thus,
$$\begin{aligned} 3^{p}<2^{b+1} & =8^{\frac{b+1}{3}}<9^{\frac{b+1}{3}}=3^{\frac{2(b+1)}{3}} \\ 2 \cdot 3^{q}<2^{b+1} & =2 \cdot 8^{\frac{b}{3}}<2 \cdot 9^{\frac{b}{3}}=2 \cdot 3^{\frac{2 b}{3}}<2 \cdot 3^{\frac{2(b+1)}{3}} \end{aligned}$$
Thus, $p, q<\frac{2(b+1)}{3}$. Since $p=b-q$ and $q=b-p$, we get $p, q>b-\frac{2(b+1)}{3}=\frac{b-2}{3}$. Therefore,
$$\frac{b-2}{3}<p, q<\frac{2(b+1)}{3}$$
Let $r=\min \{p, q\}$. Since $r>\frac{b-2}{3} \geq \frac{5}{3}$, then $r \geq 2$. Consequently, the right hand side of (2) is divisible by 9. Thus, 9 divides $2^{b+1}-1$. This is true only if $6 \mid b+1$. Since $b \geq 7$, then $b \geq 11$. Thus, we can let $b+1=6 s$ for some positive integer $s$, and we can write
$$2^{b+1}-1=2^{6 s}-1=4^{3 s}-1=\left(2^{s}-1\right)\left(2^{s}+1\right)\left(4^{2 s}+4^{s}+1\right)$$
Since $4^{s} \equiv 1 \bmod 3$, then $4^{2 s}+4^{s}+1=\left(4^{s}-1\right)^{2}+3 \cdot 4^{s}$ is always divisible by 3 but never by 9. Furthermore, at most one of $2^{s}-1$ and $2^{s}+1$ is divisible by 3, being consecutive odd numbers. Since $3^{r} \mid 2^{b+1}-1$, then either $3^{r-1} \mid 2^{s}-1$ or $3^{r-1} \mid 2^{s}+1$. From both cases, we have $3^{r-1} \leq 2^{s}+1$. Thus,
$$3^{r-1} \leq 2^{s}+1 \leq 3^{s}=3^{\frac{b+1}{6}}$$
Therefore,
$$\frac{b-2}{3}-1<r-1 \leq \frac{b+1}{6}$$
which implies $b<11$. However, there is no odd integer $b$ between 7 (included) and 11 (excluded) such that $6 \mid b+1$. There are then no solutions for this last case.

Therefore, the solutions $(x, y)$ of the given equation are $(2,1),(3,1),(3,2)$, and $(18,2)$.