I know that $$ P=IV=I^2R=\frac{V^2}{R}$$
I solved for the current using the loop rule $$I=.47$$ but to solve for the Power output of the 16V battery I could only find the power by finding the terminal voltage of the battery and using $$P=IV$$ the other two gave me incorrect answers and I'm just lost shouldn't I be able to use whichever I want?
When you solve the loop current $I$, you need the total resistance $R_{tot}$ and total voltage $V_{tot}$ to compute $I = V/R$. The total voltage is obtain by the difference between the two batteries because they compete with one another, so this is reflected in the current, which is lower than it would if it was a single battery. I guess to look at the single battery's output you need to take it's voltage $V_1$ and the total resistance $R_{tot}$.
The terminal pd is $V=\mathcal E -IR$ where the emf of the battery is $\mathcal E, \,I$ is the current and $R$ is the internal resistance.
So the power delivered to the outside circuit is $VI=\mathcal E \,I-I^2R$
The reason is that the battery itself as you represent it in your diagram does not have a resistance. It has an internal resistance $R=16~\Omega$ for which the formula $P=RI^2$ can be used but the voltage $16~V$ is pure potential such that the only valid formula is $P=VI$.
Consequently, the battery releases a power $P_V=I*16=7.52~W$ and its internal resistance dissipates $P_{R}=1.6I^2=0.35~W$. For the internal resistance all 3 expression of power should work because its a resistance.
The net output power released by the battery to the rest of the circuit can then be considered $P=P_v-P_{R}=7.52-0.35=7.71~W$.
Although I do not see the point of putting a battery $8V$ opposite to the first battery..
The relation
$$P=IV$$
is a general relation for any kind of circuit element.
The other two versions
$$P=\frac{V^2}{R}$$ and $$P=I^2R$$
are special case formulas that apply to resistors. They are derived by substituting Ohm's law into the general relationship.
Since a battery is not a resistor, you can't expect either of the formulas for a resistor to apply to it.
Similarly, you can't expect to use Ohm's law to find the voltage given the current or current given the voltage for the battery.
