This opamp is being used as a transconductance amplifier, so it doesn't work the way you normally think of an opamp working.

Assume \$V_{in}\$ is a fixed voltage 500 mV, the opamp wants to make both inputs have the same voltage so it will output enough current to cause there to be a 500 mV drop across \$R_F\$. Since the output current depends on \$V_{in}\$ and \$R_F\$, it is independent of \$R_L\$. Since \$\Delta R_L\$ has no effect on the current it's like the output impedance of the opamp is infinite.

This opamp is being used as a transconductance amplifier, so it doesn't work the way you normally think of an opamp working.

Assume \$V_{in}\$ is a fixed voltage 500 mV, the opamp wants to make both inputs have the same voltage so it will output enough current to cause there to be a 500 mV drop across \$R_F\$. Since the output current depends on \$V_{in}\$ and \$R_F\$, it is independent of \$R_L\$. Since \$\Delta R_L\$ has no effect on the current it's like the output impedance of the opamp is infinite.

Imagine a current source made with a very high voltage and a very high resistance in series with a relatively low resistance load.

^{simulate this circuit – Schematic created using CircuitLab}

With a 0 Ω load the current is 1 uA, with a 1k load it's 0.999999 uA, very little change. If you could make the voltage and resistance of the source infinite the current wouldn't change at all with a change in the load. So if the load doesn't affect the current the impedance must be infinite.