Why high voltage transmission lines?

You've begun with this:

$P_l = I^2R$

$P_l = IV$

This is correct, but the $V$ here is not the line voltage, but instead the voltage drop across the resistor under consideration. Increasing the line voltage does not increase the voltage drop.

Your diagram with a single resistance and a power station implies that the current in the line depends on that resistance and the line voltage. In reality, it does not. The resistance is just a (usually small) portion of the circuit.

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sigh I'll leave the above, but I had misread your question. I thought the diagram was oversimplified, but that's actually the situation you were asking about.

Because there is only one source of loss (the resistor), then power lost through is is equal to power generated. $P_l = P_g$. You cannot now state that $P_g$ is a constant, the resistance is a constant, and voltage is variable.

This is great, as lowering the current means raising the voltage,

It does for constant power. But you want to have constant resistance instead. You cannot have both while varying voltage.

High voltage, because:

The voltage that a power plant pushes out, is consumed by wires (loss) and a machine (desired). The smaller voltage the wires consume, the smaller the loss.

In order to minimize loss, you have to minimize the voltage share of the lines.

Voltage consumed by the wires is $V=RI$ ($R$ is resistance of the lines). As you can see, the lower the current ($I$), the lower gets the voltage consumed by the wires.

As $P = VI$, to get low current, you have to get high voltage to get the total power of the system to remain. In this latter equation V is the voltage pushed by the powerplant.