Can I charge a 16V 16.6F supercapacitor with an 18V solar panel?
To avoid damage to the supercapacitor, it must not be given a voltage above its rating. A 18V solar panel, on a sunny day, will generate 18V. The diode (if silicon) will drop 0.6-0.7V depending upon the diode and current. That leaves 17.3V. While this might not damage the capacitor, it very well might. It seems best to avoid that possibility.
The circuit example you give contains a shunt over-voltage protection element. A shunt OVP will protect a circuit from over voltage by shunting away current. Thus, on a sunny day, when the solar panel is generating at 18V, much or most of the current generated may be shunted, with great waste of power.
A better solution is to use a series over-voltage protection circuit, rather than a shunt. What this does is add resistance to your circuit to ensure that the output voltage does not exceed a given level. The advantage is more power is available to your load than with the shunt. (The disadvantage, is that even when there is no over-voltage condition, there is some a voltage drop, and consequent loss of power).
Some low drop-out regulators (LDO's) will function quite nicely for this purpose. Although their normal mode of operation is to provide a regulated output given an input sufficiently above their rated output, appropriate devices will provide output even when their input is below the minimum required for regulation, them simply provide an unregulated output, at a voltage slightly lower than the input.
An example circuit:



simulate this circuit &ndash; Schematic created using CircuitLab