Given the equation $x^{2}+y^{2}-4x-2y-4=0$, we can rewrite it by completing the squares for both $x$ and $y$:

$$x^{2}-4x+y^{2}-2y=4$$

$$\Rightarrow (x^{2}-4x+4)+(y^{2}-2y+1)=4+4+1$$

$$\Rightarrow (x-2)^{2}+(y-1)^{2}=9$$

This represents a circle with center at $(2,1)$ and radius $3$.

Now, let $z=x-y$, which implies $x=y+z$. Substituting $x=y+z$ into the equation of the circle, we get:

$$(y+z-2)^{2}+(y-1)^{2}=9$$

This equation represents the condition for the intersection of the line $y=x-z$ with the circle.

To find the maximum value of $z=x-y$, we consider the distance from the center of the circle to the line $y=x-z$, which can be expressed as:

$$\frac{|2-1-z|}{\sqrt{1^2+1^2}} \leqslant 3$$

Simplifying the inequality:

$$\frac{|1-z|}{\sqrt{2}} \leqslant 3$$

$$|1-z| \leqslant 3\sqrt{2}$$

This inequality can be split into two cases:

1. $1-z \leqslant 3\sqrt{2}$, which gives $z \geqslant 1-3\sqrt{2}$

2. $1-z \geqslant -3\sqrt{2}$, which simplifies to $z \leqslant 1+3\sqrt{2}$

Combining these two cases, we find that:

$$1-3\sqrt{2} \leqslant z \leqslant 1+3\sqrt{2}$$

Therefore, the maximum value of $x-y$ is $\boxed{1+3\sqrt{2}}$, which corresponds to option C.