Common emitter: maximize effective voltage gain
The gain of the common emitter amplifier is not determined by the collector resistance, but rather by the Q point, where the collector resistance takes into account the load. What happens is that if you increase the collector resistance, the base-emitter resistance \$r_\mathrm{e}\$ increases to compensate for the same Q point.

Under the assumptions that the collector current is equal to the emitter current, the emitter resistor is perfectly bypassed by the emitter capacitor, and the transistor is in small signal operation, the gain is
$$|A_\mathrm{v}| = \frac{R_\mathrm{c}}{r_\mathrm{e}} = \frac{(V_\mathrm{cc} - V_\mathrm{Q}) / I_\mathrm{Q}}{k_\mathrm{B}T / (q I_\mathrm{Q})} = \frac{q}{k_\mathrm{B}T} (V_\mathrm{cc} - V_\mathrm{Q}) \approx 40(V_\mathrm{cc} - V_{Q}),$$
where \$I_\mathrm{Q}\$ is the quiescent current, \$V_\mathrm{Q}\$ is the quiescent voltage across the collector, \$r_\mathrm{e}\$ is the base emitter resistance, \$k_\mathrm{B}\$ is the Boltzmann constant, \$T \approx 290\ \mathrm{K}\$ is the absolute temperature, \$q\$ is the electron charge, and \$V_\mathrm{cc}\$ is the positive power rail.

As you can see the gain does not depend on the collector resistor. To maximize the gain one needs to bring the Q point as low as possible. However you must ensure that the Q point is far enough from the emitter voltage in order to allow the transistor to swing down. You must also take into account the voltage across the emitter resistor needed for thermal stability.