Concept:
Coordinate Geometry - Intersection of a Line and a Circle.
Step 1: Find the center and radius of the given circle.
The equation of the circle is $x^{2}+y^{2}-2x-4y=0$. We can rewrite this in standard form by completing the square: $(x^2-2x+1) + (y^2-4y+4) = 1 + 4$.
This simplifies to $(x-1)^2 + (y-2)^2 = 5$. Thus, the center $C$ is $(1, 2)$ and the radius $r$ is $\sqrt{5}$.
Step 2: State the condition for a line to intersect a circle at two distinct points.
For a line to intersect a circle at two distinct points, the perpendicular distance ($d$) from the center of the circle to the line must be strictly less than the radius of the circle ($d<r$).
Step 3: Calculate the perpendicular distance from the center to the line.
The line equation is $x - 2y - m = 0$. The perpendicular distance $d$ from the center $(1, 2)$ to this line is given by the formula $d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}$. Substituting our values gives: $d = \frac{|1(1) - 2(2) - m|}{\sqrt{1^2 + (-2)^2}} = \frac{|1 - 4 - m|}{\sqrt{1 + 4}} = \frac{|-3 - m|}{\sqrt{5}} = \frac{|m + 3|}{\sqrt{5}}$.
Step 4: Apply the condition and solve the inequality for $m$.
Setting $d<r$, we get $\frac{|m + 3|}{\sqrt{5}}<\sqrt{5}$. Multiplying both sides by $\sqrt{5}$ yields $|m + 3|<5$. Removing the absolute value gives the compound inequality $-5<m + 3<5$. Subtracting 3 from all parts gives $-8<m<2$.
Step 5: Count the integer values of $m$.
The problem states that $m \in \mathbb{Z}$ (m is an integer). The integers strictly between -8 and 2 are: -7, -6, -5, -4, -3, -2, -1, 0, 1. Counting these yields a total of 9 possible values.
$$
\therefore \text{The number of possible values of } m \text{ is } 9.
$$