14. From the conditions, we know
$$\begin{aligned} b-a & =(\sqrt{x+1}-\sqrt{x-1})+(\sqrt{y+1}-\sqrt{y-1}) \\ & =\frac{2}{\sqrt{x+1}+\sqrt{x-1}}+\frac{2}{\sqrt{y+1}+\sqrt{y-1}} \\ & 2-\frac{2}{\sqrt{2}}=2-\sqrt{2} \end{aligned}$$

Therefore,
$$\sqrt{x+1}+\sqrt{x-1}<\frac{2}{2-\sqrt{2}}=2+\sqrt{2}$$

Similarly,
$$\sqrt{y+1}+\sqrt{y-1}<2+\sqrt{2}$$

This indicates
$$a+b=(\sqrt{x+1}+\sqrt{x-1})+(\sqrt{y+1}+\sqrt{y-1})<4+2 \sqrt{2},$$

Thus,
$$a+b \leqslant 6 \text {. }$$

Combining $b-a=2$ and the fact that $b-a$ and $b+a$ have the same parity, we have
$$\begin{array}{c} (b-a, b+a)=(2,4),(2,6) \\ (a, b)=(1,3),(2,4) \end{array}$$

Solving each case, we find that only $(a, b)=(1,3)$ has a solution, which is
$$x=y=\frac{5}{4} .$$