Headphone output is anemic for a desktop product:
If voltage supply is the limitting factor, this gives an output impedance of around 100 Ohm.
Since by Ohms law V = sqrt(P*Z), where Z is impedance and by voltage division in parallel circuits, V = Z/(Z + Z_O)*V_O, where V is the voltage across the load measured, V_O is the output voltage and Z_O is the output impedance,
sqrt(P*Z) = Z/(Z + Z_O)*V_O.
Solving for V_O gives
V_O = sqrt(P*Z)*(Z + Z_O)/Z = sqrt(P*Z)*(1 + Z_O/Z) = sqrt(P*Z) + Z_O*sqrt(P/Z).
If voltage supply is the limitting factor, the output voltage at the measured powers, P_1 and P_2, for two different load impedances, Z_1 and Z_2, is the same. Thus,
sqrt(P_1*Z_1) + Z_O*sqrt(P_1/Z_1) = sqrt(P_2*Z_2) + Z_O*sqrt(P_2/Z_2)
which gives
Z_O*(sqrt(P_1/Z_1) - sqrt(P_2/Z_2)) = sqrt(P_2*Z_2) - sqrt(P_1*Z_1)
and thus the output impedance is
Z_O = (sqrt(P_2*Z_2) - sqrt(P_1*Z_1))/(sqrt(P_1/Z_1) - sqrt(P_2/Z_2)).
Note that the unit of power is irrelevant as long as it is consistent.
Using the measured powers at 32 and 300 Ohm, this gives,
Z_O = (sqrt(31*300) - sqrt(29*32))/(sqrt(29/32) - sqrt(31/300)) = 104.63
Assuming instead that the powers are the same,
Z_O = (sqrt(300) - sqrt(32))/(sqrt(1/32) - sqrt(1/300)) = 97.98