Optocoupler to switch 24V using Arduino
the available output current with an optocoupler is limited to small values.

You cannot directly switch a relatively high-current load with an optocoupler. But it can be used as a pre-amplifier (or pre-driver) for a switching transistor.

How would I redesign my circuit to include an optocoupler and which optocoupler whould be necessary, considering I'm not sending data but only an on-off signal that doesn't need to switch back and forth too frequently?

Recommending specific parts is rated as off-topic here. Nevertheless, I'll recommend you to use any 817 series optocouplers as they are too common and very easy to find anywhere in the world. The ones with relatively higher CTR (Current transfer ratio) would be better, so HCPL817 (any CTR rank starting from B) can be a good option.
CTR for an optocoupler is simply a ratio of collector current to LED current. You can think of it as something like hFE of BJTs.
As for using it to switch a load, here's a circuit you can use:



simulate this circuit &ndash; Schematic created using CircuitLab
I selected BC547 as the load switching transistor because it has relatively high hFE (I personally tend to use 2N390x for relatively faster switching applications). Since the BJT should operate in saturation mode here, a base current of one-twentieth to one-tenth of the load current is sufficient (this is a practical tip for small-signal BJTs). So, for \$\mathrm{I_C=175mA}\$, the minimum base current could be \$\mathrm{I_B=I_C/15=12mA}\$ is quite sufficient.
\$\mathrm{I_B}\$ of the BJT flows through the optocoupler's transistor. Assuming a 1V drop across the optocoupler's transistor at 12mA, the current limiter resistor for BJT's base could be \$\mathrm{R2 = (24V - 1V)/12mA = 1k9}\$. The nearest standard value is 1k8. R5 is there to guarantee the off state of BJT (i.e. to prevent any base-emitter instability which can cause an accidentally turn-on).
Assuming the minimum CTR of the optocoupler is 100%, the required minimum LED current is 12mA. The forward voltage drop of optocoupler LED is around 1.1V at 12mA, so the current limiter resistor is \$\mathrm{R1 = (5V - 1.1V)/12mA \approx 3k3}\$.