1. Let $|z|=r \quad (r>0), \qquad z=re^{i\theta}.$ Since the expression contains $\dfrac{4}{z}$, we must have $z\neq 0$.

2. Rewrite the given condition: $\left|z-\frac{4}{z}\right|=2.$ Substitute $z=re^{i\theta}$: $z-\frac{4}{z}=re^{i\theta}-\frac{4}{r}e^{-i\theta}.$

3. Take modulus squared: $\left|re^{i\theta}-\frac{4}{r}e^{-i\theta}\right|^2=4.$ Using $|a-b|^2=|a|^2+|b|^2-2\Re(a\overline b)$, or direct expansion, $r^2+\frac{16}{r^2}-8\cos 2\theta = 4.$ So, $8\cos 2\theta = r^2+\frac{16}{r^2}-4.$ Hence $\cos 2\theta = \frac{r^2+\frac{16}{r^2}-4}{8}.$

4. For some complex number $z$ to exist, we need a real $\theta$ such that $-1\le \cos 2\theta \le 1.$ Therefore, $-1\le \frac{r^2+\frac{16}{r^2}-4}{8} \le 1.$ To maximize $r$, it is enough to use the upper bound: $\frac{r^2+\frac{16}{r^2}-4}{8} \le 1.$ This gives $r^2+\frac{16}{r^2} \le 12.$ Multiply by $r^2$: $r^4-12r^2+16\le 0.$ Let $x=r^2$. Then $x^2-12x+16\le 0.$ Solve the quadratic: $x=\frac{12\pm\sqrt{144-64}}{2}=\frac{12\pm\sqrt{80}}{2}=6\pm 2\sqrt5.$ So, $6-2\sqrt5\le r^2\le 6+2\sqrt5.$ Hence the maximum possible value of $r=|z|$ is $r_{\max}=\sqrt{6+2\sqrt5}.$ Now, $(\sqrt5+1)^2=5+1+2\sqrt5=6+2\sqrt5,$ therefore $\sqrt{6+2\sqrt5}=\sqrt5+1.$

5. Thus, $\boxed{\max |z|=\sqrt5+1}.$

6. Option check:

• A: $\sqrt5+1$ ✅
• B: $2$ ❌
• C: $2+\sqrt2$ ❌
• D: $\sqrt3+1$ ❌

So the correct option is A.