Let $|z|=r \quad (r>0), \qquad z=re^{i\theta}.$

Then $\frac{1}{z}=\frac{1}{r}e^{-i\theta}.$ So $z-\frac{1}{z}=re^{i\theta}-\frac{1}{r}e^{-i\theta}.$

Now compute its modulus squared: [ \left|z-\frac{1}{z}\right|^2 =\left(re^{i\theta}-\frac{1}{r}e^{-i\theta}\right) \left(re^{-i\theta}-\frac{1}{r}e^{i\theta}\right). ]

Expanding, [ =r^2+\frac{1}{r^2}-e^{2i\theta}-e^{-2i\theta}. ]

Since $e^{2i\theta}+e^{-2i\theta}=2\cos 2\theta,$ we get $\left|z-\frac{1}{z}\right|^2=r^2+\frac{1}{r^2}-2\cos 2\theta.$

Given $\left|z-\frac{1}{z}\right|=2,$ hence $r^2+\frac{1}{r^2}-2\cos 2\theta=4.$

Therefore, $2\cos 2\theta=r^2+\frac{1}{r^2}-4,$ so $\cos 2\theta=\frac{1}{2}\left(r^2+\frac{1}{r^2}-4\right).$

For some complex number $z$ to exist, we must have $-1\le \cos 2\theta \le 1.$ Hence, $-2 \le r^2+\frac{1}{r^2}-4 \le 2.$

Adding 4 throughout, $2 \le r^2+\frac{1}{r^2} \le 6.$

To maximize $r$, use the upper bound: $r^2+\frac{1}{r^2}\le 6.$ Let $x=r^2>0.$ Then $x+\frac{1}{x}\le 6.$

Multiply by $x$: $x^2-6x+1\le 0.$

Solve the quadratic equation $x^2-6x+1=0.$ Its roots are $x=\frac{6\pm\sqrt{36-4}}{2}=\frac{6\pm\sqrt{32}}{2}=3\pm 2\sqrt{2}.$

Therefore, $3-2\sqrt{2} \le x \le 3+2\sqrt{2}.$

So the maximum value of $x=r^2$ is $r^2_{\max}=3+2\sqrt{2}=(\sqrt{2}+1)^2.$

Hence, $r_{\max}=|z|_{\max}=\sqrt{3+2\sqrt{2}}=\sqrt{2}+1.$

Therefore, the maximum value of $|z|$ is $\boxed{\sqrt{2}+1}.$