2. Find all pairs of non-negative integers $(m, n)$ such that $m^{2}+2 \times 3^{n}=m\left(2^{n+1}-1\right)$.
2. For a fixed $n$, the original equation is a quadratic equation in $m$.
When $n=0,1,2$, the discriminant $\Delta \frac{n-2}{3}$.
In particular, $h>1$.
From equation (1), we know
$$
3^{k}\left|\left(2^{n+1}-1\right) \Rightarrow 9\right|\left(2^{n+1}-1\right) \text {. }
$$

Since the order of 2 modulo 9 is 6, it follows that $6 \mid (n+1)$.
Let $n+1=6r\left(r \in \mathbf{N}_{+}\right)$. Then
$$
\begin{array}{l}
2^{n+1}-1=4^{3r}-1 \\
=\left(4^{2r}+4^{r}+1\right)\left(2^{r}-1\right)\left(2^{r}+1\right) .
\end{array}
$$

The factor of equation (3)
$$
4^{2r}+4^{r}+1=\left(4^{r}-1\right)^{2}+3 \times 4^{r}
$$

is divisible by 3 but not by 9, and $2^{r}-1$ and $2^{r}+1$ are coprime, so
$3^{h-1} \mid\left(2^{r}-1\right)$ or $3^{h-1} \mid\left(2^{r}+1\right)$.
Thus, $3^{h-1} \leqslant 2^{r}+1 \leqslant 3^{r}=3^{\frac{n+1}{6}}$.
Therefore, $\frac{n-2}{3}-1<h-1 \leqslant \frac{n+1}{6}$.
So, $6 \leqslant n<11 \Rightarrow 6 \mid (n+1)$. Contradiction.
In summary, the solutions that satisfy the conditions are
$$
(m, n)=(6,3),(9,3),(9,5),(54,5) .
$$