Can 720! be written as the difference of two positive integer powers of 3? Does the equation:
$$3^x-3^y=720!$$
have any positive integer solution?

If $$720!=3^x-3^y=3^y\left(3^{x-y}-1\right)$$ and since the power of $3$ dividing $720!$ is $$\left\lfloor\frac{720}{3}\right\rfloor+\left\lfloor\frac{720}{9}\right\rfloor+\left\lfloor\frac{720}{27}\right\rfloor+\left\lfloor\frac{720}{81}\right\rfloor+\left\lfloor\frac{720}{243}\right\rfloor=240+80+26+8+2=356\text{,}$$ it would have to be that $y=356$. 
So it remains to see if $3^x-3^{356}=720!$ has an integer solution in $x$. 

Side note: we can get a good approximation to $\log_3(720!)$ using Stirling's formula with one more term than is typically used: $$\log_3(720!)\approx\frac{1}{\ln3}\left(720\ln(720)-720 +\frac{1}{2}\ln(2\pi\cdot720)\right)\approx3660.3\ldots$$ Since the terms in Stirling's formula are alternating after this, we can deduce that this is correct to the tenths place. Values of $\ln(3)$, $\ln(720)$, and $\ln(2\pi)$ are easy to calculate by hand to decent precision if needed. 

$720!$ is a lot bigger than $3^{356}$. Since  $\log_3(720!)\approx3660.3$, in base $3$, $720!$ has $3661$ digits (trigits?), where as $3^{356}$ just has $357$. So $\log_3(720!+3^{356})$ and $\log_3(720!)$ must be very close together. Since the latter is $\approx3660.3$ though, it's not possible for the former to be an integer.
More formally, $$\log_3(720!)<\log_3(720!+3^{356})=x=\log_3(720!)+\log_3\mathopen{}\left(1+\frac{3^{356}}{720!}\right)\mathclose{}<\log_3(720!)+\frac{1}{\ln(3)}\frac{3^{356}}{720!}$$
$$3660.3\ldots<x<3660.3\ldots$$
and there is no integer $x$ between the values on the two ends.
