Solution:
(Ⅰ) From the given conditions, we have: $\begin{cases} m^2 = 4p \\ \sqrt{4 + m^2} = 2\sqrt{3} \end{cases}$,
Solving this, we get $p = 2$,
Therefore, the equation of the parabola $E$ is $y^2 = 4x$.
(Ⅱ) Since point $A(2, m)$ lies on the parabola $E: y^2 = 4x$,
we have $m = \pm 2\sqrt{2}$,
Considering the symmetry of the parabola, let's assume $A(2, 2\sqrt{2})$.
From $A(2, 2\sqrt{2})$ and $F(1, 0)$, we get the equation of line $AF$ as $y = 2\sqrt{2}(x - 1)$.
From $\begin{cases} y = 2\sqrt{2}(x - 1) \\ y^2 = 4x \end{cases}$, we get $2x^2 - 5x + 2 = 0$,
Solving this, we find $x = 2$ or $x = \frac{1}{2}$, thus $B\left(\frac{1}{2}, -\sqrt{2}\right)$.
Also, for $G(-1, 0)$,
we have $k_{GA} = \frac{2\sqrt{2} - 0}{2 - (-1)} = \frac{2\sqrt{2}}{3}$, $k_{GB} = \frac{-\sqrt{2} - 0}{\frac{1}{2} - (-1)} = -\frac{2\sqrt{2}}{3}$,
Therefore, $k_{GA} + k_{GB} = 0$, hence $\angle AGF = \angle BGF$,
This indicates that the distance from point F to lines GA and GB are equal,
Thus, a circle with center at point F and tangent to line GA must also be tangent to line GB.

The final answers are:
(Ⅰ) The equation of the parabola $E$ is $\boxed{y^2 = 4x}$.
(Ⅱ) It is proven that a circle with center at point F and tangent to line GA must also be tangent to line GB.