Integer exponent equation 
Show that $$(2^a-1)(2^b-1)=2^{2^c}+1$$ doesn't have solution in positive integers $a$, $b$, and $c$.

After expansion I got
$$2^{a+b}-2^a-2^b=2^{2^c}\,.$$
Any hint will be appreciated.
 A: $$2^{a+b}=2^a+2^b+2^{2^c}$$
Case $1$:
If $a=b$, then $$2^{2a}=2^{a+1}+2^{2^c}$$
If $a+1$ and $2^c$ are distinct, then surely, the hamming weight on the RHS is $2$ but the hamming weight on the left is $1$.
Hence we must have $a+1=2^c$
$$2^{2a}=2^{a+2}$$
Hence $2a=a+2$, hence $a=2$, but since $a+1=2^c$, we have $3=2^c$ which is a contradiction.
Case $2: a \ne b$, if $2^c$ is not equal to $a$ or $b$, then the hamming weight on the right is $3$ but the hamming weight on the left is $1$. Also, let $a> b$. In order to have the hamming weight of $2^a+2^b$ to drop to $1$ upon adding $2^{2^c}$. We need $a=b+1$ and $b=2^c$.
$$2^{2b+1}=2^{b+1}+2^b+2^b=2^{b+2}$$
$$2b+1=b+2$$
$$b=1$$
Hence $c=0$.
Hence, we do not have positive solution.
A: Hint: Using $2^{a+b}-2^a-2^b=2^{2^c}$, show if $a \neq b$, say WLOG $a \lt b$, then there's only $a$ factors of $2$ on the LHS and, apart from the case of $a = 1$ and $b = 2$ (which doesn't work since $c = 0$ is not allowed), an odd factor $\gt 1$. This means you must have $a = b$. However, then show this doesn't work either.
