Solution:
$(1)$ Let $C(x,y)$, then according to the problem, we have $k_{AC} \cdot k_{BC} = - \dfrac {3}{4}$,
Since $A(-2,0)$, $B(2,0)$, we get $\dfrac {y}{x+2} \cdot \dfrac {y}{x-2} = - \dfrac {3}{4} (y \neq 0)$,
After rearranging, we get $\dfrac {x^{2}}{4} + \dfrac {y^{2}}{3} = 1 (y \neq 0)$,
$\therefore$ The trajectory equation of point $C$ is $\dfrac {y}{x+2} \cdot \dfrac {y}{x-2} = - \dfrac {3}{4} (y \neq 0)$.
Proof:
$(2)$ Proof method $1$: Let line $l: y=kx+m$, and it intersects with $3x^{2}+4y^{2}=12$,
By combining them, we get $3x^{2}+4(kx+m)^{2}=12$, which simplifies to $(3+4k^{2})x^{2}+8kmx+4m^{2}-12=0$,
According to the problem, $\Delta = (8km)^{2}-4(3+4k^{2})(4m^{2}-12)=0$, thus $3+4k^{2}=m^{2}$,
Let line $l$ intersect the trajectory of moving point $C$ at points $(x_{1},y_{1})$, $(x_{2},y_{2})$,
$\therefore x_{1}+x_{2}= \dfrac {-8km}{3+4k^{2}}$, we get $x_{1}=x_{2}= \dfrac {-4km}{3+4k^{2}}$,
$\therefore P( \dfrac {-4km}{3+4k^{2}}, \dfrac {3m}{3+4k^{2}})$, and since $3+4k^{2}=m^{2}$, we get $P( \dfrac {-4k}{m}, \dfrac {3}{m})$, also $Q(4,4k+m)$,
Given $F(1,0)$, then $\overrightarrow{FP} \cdot \overrightarrow{FQ}=(- \dfrac {4k}{m}-1, \dfrac {3}{m}) \cdot (3,4k+m)=0.$ It is known that $\overrightarrow{FP} \perp \overrightarrow{FQ}$,
thus $\angle PFQ=90^{\circ}$.
Proof method $2$: Let $P(x_{0},y_{0})$, then the tangent line $PQ$ of curve $C$ at point $P$ is: $\dfrac {x_{0}x}{4}+ \dfrac {y_{0}y}{3}=1$,
By setting $x=4$, we get $Q(4, \dfrac {3-3x_{0}}{y_{0}})$, and given $F(1,0)$,
$\therefore \overrightarrow{FP} \cdot \overrightarrow{FQ}=(x_{0}-1,y_{0}) \cdot (3, \dfrac {3-3x_{0}}{y_{0}})=0.$ It is known that $\overrightarrow{FP} \perp \overrightarrow{FQ}$,
thus $\angle PFQ=90^{\circ}$.

Therefore, the final answers are:
$(1)$ The trajectory equation of point $C$ is $\boxed{\dfrac {x^{2}}{4} + \dfrac {y^{2}}{3} = 1 (y \neq 0)}$.
$(2)$ It is proven that $\boxed{\angle PFQ=90^{\circ}}$.