Find integer a,b > 1 such that $2^a + 3^b = 2^{a+b} +1$ I would like to know if it is possible to find an integer solution to
$2^a + 3^b = 2^{a+b} +1$
with $a,b > 1$

We have 
$$2^a(2^b - 1) = (3-1)(1 + 3 + 3^2 + \ldots + 3^{b-1}),$$ so
$$2^{a-1}(2^b - 1) = (1 + 3 + 3^2 + \ldots + 3^{b-1}).$$ 
Considering the second equation mod $3$ we thus see that $2^{a-1}$ and $2^b - 1$ aren't zero. As $2^b$ isn't zero mod $3$ either, it must be $2$ mod $3$, i.e. $b$ must be odd. 
But this implies that $1 + 3 + 3^2 + \ldots + 3^{b-1}$ is $2$ mod $4$ and hence (from the first equality in this answer) that $a = 2$. 
But there are no solutions with $a = 2$, since then we have $4 - 1 + 3^b = 2^{a+b}$ and the two sides are different modulo $3$.
