Let $(m, n)$ be a pair of positive integers satisfying (14), then $n>m$.
Notice
$$\begin{aligned} {[(m+3) \sqrt{2}] } & =[m \sqrt{2}+3 \sqrt{2}] \geqslant[m \sqrt{2}]+[3 \sqrt{2}] \\ & =[m \sqrt{2}+4] \end{aligned}$$

Therefore, $n=m+1$ or $n=m+2$.
If $n=m+1$, since $\frac{1}{\sqrt{2}}+\frac{1}{2+\sqrt{2}}=1$, by the Beatty-Rayleigh theorem, we know that $\{[\sqrt{2} m]\}$ and $\{[(2+\sqrt{2}) h]\}$ are complementary sequences, and at this time
$$[n \sqrt{2}]=[(m+1) \sqrt{2}]=[2+m \sqrt{2}]=2+[m \sqrt{2}],$$

Thus, there is exactly one integer $[(2+\sqrt{2}) h]$ between $[m \sqrt{2}]$ and $[(m+1) \sqrt{2}]$, i.e.,
$$m \sqrt{2}<(2+\sqrt{2}) h<(m+1) \sqrt{2}$$

Hence, $m<(\sqrt{2}+1) h<m+1$, i.e.,
$$m=[(\sqrt{2}+1) h]$$

Reversing the above process, we know that
$$(m, n)=([(\sqrt{2}+1) h],[(\sqrt{2}+1) h]+1)$$

is a solution to (14). Therefore, the positive integer solutions to (14) in this case are

If $n=m+2$, similarly, we know $[(m+2) \sqrt{2}]=[2+m \sqrt{2}]=2+[m \sqrt{2}]$, so $[m \sqrt{2}],[(m+1) \sqrt{2}]$, and $[(m+2) \sqrt{2}]$ are three consecutive positive integers. Therefore, there exists $h \in \mathbf{N}^{*}$, such that
$$[(2+\sqrt{2})(h+1)]-[(2+\sqrt{2}) h] \geqslant 4$$

However,
$$[(2+\sqrt{2})(h+1)]-[(2+\sqrt{2}) h] \leqslant 1+[2+\sqrt{2}]=4$$

Thus, there exists $h \in \mathbf{N}^{*}$, such that
$$\left\{\begin{array}{l} {[(m-1) \sqrt{2}]<[(2+\sqrt{2}) h]<[m \sqrt{2}]} \\ {[(2+\sqrt{2})(h+1)]=[(2+\sqrt{2}) h]+4} \end{array}\right.$$

From (15), we know $m=[(1+\sqrt{2}) h]+1$, and (16) is $[\sqrt{2}(h+1)]=[\sqrt{2} h+2]$. Therefore, using the discussion of the first case ($h$ is equivalent to $m$), we know there exists $k \in \mathbf{N}^{*}$, such that $h=[(1+\sqrt{2}) k]$. Reversing the above process, we know that the positive integer solutions to (14) in this case are
$$\left\{\begin{array}{l} m=[(1+\sqrt{2}) h]+1=3 k+2[\sqrt{2} k], \\ n=3 k+2+2[\sqrt{2} k] \end{array}\right.$$

In the last step, we use
$$\begin{aligned} m & =[(1+\sqrt{2})[(1+\sqrt{2}) k]]+1 \\ & =[(3+2 \sqrt{2}) k+1-(1+\sqrt{2}) \alpha] \\ & =[3 k+2[\sqrt{2} k]+1-(\sqrt{2}-1) \alpha] \\ & =3 k+2[\sqrt{2} k] \end{aligned}$$

Here $\alpha=\{(1+\sqrt{2}) k\}=\{\sqrt{2} k\}$.