Zener diode behaviour at operating limit
You should base your analysis on  \$I_{ZT}\$ instead of \$I_{ZK}\$. The 5mA current is the one that guarantees that the voltage is going to be under regulation.


  
    Does the zener voltage depends on the current?
  


Yes, that's why they give you an impedance (e.g 89 Ohms max @\$I_{ZT}\$) when working as a regulator. So if the current changes, so does the zener voltage.

\$I_{ZK}\$ and \$Z_{ZK}\$ are the current and impedance they get near the breakdown region (the knee in the IV curve).

In order to be certain that the zener will work as a voltage regulator, you have to ensure there is always 5mA (the test current they use) running through it.

Your minimum sustained voltage is 15V, your zener voltage is 5V, and you want 5mA flowing to the zener at all times to keep the voltage regulated.

Then,

$$ R_s = \dfrac{V_s-V_z}{I_t}$$

Where \$I_t\$ is the total current from the power supply (zener current + current drawn by the pin or load). So if the pin draws an additional 10mA for exaple, your total current is 15mA.

$$ R_s = \dfrac{15\text{V}-5\text{V}}{15\text{mA}}=670\Omega$$

\$R_s\$ is a resistor between your source and the zener diode.

If the voltage goes any higher, say 30V as you mentioned, then there will be more current going through the zener, but as long as you don't go over the power dissipation limit of the diode, you should be ok. So if the voltage increases to 30V, the total current will be around 37mA with the chosen \$R_s\$. If you had no load, the zener will dissipate a max power of \$P_z\$=(37mA)(5V) = 0.187Watts, so choose accordingly.