Electronic – Optocoupler driving 300mA continously (or use darlington topology?)

It's impossible to know the behaviour at a current that is not the specified operating point if there's no datasheet containing an "intensity over I_f" graph – if this is of concern to you, buy LEDs that come with such a datasheet.

Yet another thing I'd like to point out that you don't say anything about your 300mA supply – it might look that an LED that is more efficient at 300mA would be the power-efficient choice, but if it happens to have a lower forward voltage at that current (is there at least an I/U graph?), you'll just "burn" that additional energy (voltage = energy per charge) in a linear regulator.

Now, it's really impossible to tell what kind of regulator you have picked there – yes, this looks a bit like the mains voltage gets rectified on the input side, and then chopped up and downconverted by the transformer type thingie there, but I can't tell the least about how exact the current regulation is (but I do have a hunch that without any secondary side sensing, it's not going to be that exact) or how efficient the conversion. So: if you're at all concerned about operational specifics of your LED system, this is not the supply of your choice.

Generally, people will sell their LEDs at the point at which they produce the maximal amount of light without thermal damage (or without exhaustion of recombination opportunities in the semiconductor), and that will be pretty close to the point of maximum efficiency. So, if you want to drive an LED specified for 700mA at 3.3V with less than its maximum power, you typically use PWM to switch it off and on with a configurable duty cycle, driving it at an "efficient" 700mA when on.

The LEDs are designed for a constant current supply which produces whatever voltage is necessary so reducing the current would be effective in dimming the LEDS.

Rohat calculated that the supply output would be a constant 700mA and for 3W per light you need 86mA each. Putting 8 lights in parallel would give 700/8=87.5mA per light which is about right, but that assumes they will share the current equally. The problem is that LEDs have a fairly sharp turn-on so whichever bulb has the lowest forward voltage will take almost all the current and die, then the next lowest would blow and so on. Unfortunately, whichever takes the highest current will heat up most and that will probably drop the voltage futher giving thermal runaway.

What you would need to do is add something to equalise the currents, a series resistor as shown would do but I suspect the value needs to be higher than shown in the circuit, at least the maximum forward voltage variation divided by the current variation you can tolerate. As the LEDs can take up to 350mA, that is fairly easy but you probably want to keep the brightness of the bulbs fairly similar so for example 10% brightness variation implies no more than 8.6mA difference between the bulbs. Higher values of resistor will maintain more even brightness but will increase heat dissipated in the resistors. The advantage is that including 8 resistors is a lot cheaper than separate drivers.

P.S. You probably realise that the supply voltage need only be 35V plus the resistor drop so a 144V-capable supply is much more than you need, it is really intended for driving multiple bulbs in series so that all get exactly the same current and hence brightness.