Example 6. Find all positive integers $m, n$ such that $2^{m}+3^{n}$ is a perfect square.
Proof: Let $2^{m}+3^{v}=x^{2}$, then it is easy to prove $2+x, 3+x$. At this point, $\quad x^{2}=(3 a \pm 1)^{2} \equiv 1(\bmod 3)$.
Therefore, it must be that $2^{m}=1(\bmod 3)$, hence $2 \mid m$. Let $m=2 s$, then from $2^{2 s}=x^{2}-3^{\mathrm{n}}$ we know
$$
x^{2}-3^{\mathrm{n}} \equiv 0(\bmod 4),
$$

Thus, $\quad x^{2}-(-1)^{n} \equiv 0(\bmod 4)$.
Noting that $2+x$, hence $x^{2} \equiv 1(\bmod 4)$. Therefore, from the above equation, it must be that $2 \mid n$. Let $n=2 t$, then
$$
2^{25}=x^{2}-3^{2 \mathrm{t}}=\left(x-3^{t}\right)\left(x+3^{t}\right) \text {. }
$$

Since $x$ is a prime number, $x \pm 3^{2}$ are both composite numbers. From (13), we know
$$
x-3^{t}=2^{\alpha}, x+3^{t}=2^{\beta},
$$

where $\alpha, \beta \geqslant 1$. Clearly, $2^{\beta}>2^{\alpha}$, so $\beta>\alpha$. Therefore, if $\alpha \geqslant 2$, then $\beta \geqslant 3$, thus $4 \mid\left(2^{\alpha}+2^{\beta}\right)$. But from (14) we know
$$
2^{\alpha}+2^{\beta}=2 x \text {, }
$$

Thus, $4 \mid 2 x$. This contradicts $2+x$, so it must be that $\alpha=1$. Therefore,
$$
x-3^{t}=2, x+3^{t}=2^{2 s-1} \text {. }
$$

Eliminating $x$ from the above equations, we get
$$
3^{t}+1=2^{2 t-2} \text {. }
$$

Clearly, $2 s-2>0$, so $s \geqslant 2$. Thus, from (15) we know
$$
3^{t}+1 \equiv 0(\bmod 4) \text {. }
$$

Thus, $2+t$, and expanding the left side of (15), we get
$$
2^{2 s-2}=(3+1)\left(3^{t-1}-3^{t-2}+\cdots-3\right.
$$
The second factor on the right side of (16) is the algebraic sum of $t$ odd numbers, and $t$ is odd, so the sum is odd. If $t>1$, then this algebraic sum is an odd number not less than 3, which contradicts the right side of (16) being a power of 2. Therefore, it must be that $t=1$, thus $s=2$, $m=4, n=2$. That is, only $2^{4}+3^{2}$ is a perfect square.

Finally, it should be noted that the methods for determining whether a number is a perfect square are not limited to the three methods discussed in this article.