Prove that there are only 3 integer solutions to $2^a+3 ^b=5^c$ I have figured out that $a \pmod{2} = b\pmod{2} =c \pmod{2} $ by making a few modulo tables and using that both sides of the equation must be divisible by $5$, and I have found the three solutions, I'm just at a loss as to how to prove that those are the only three. I struggle specifically with how to move from something being divisible by $5$ to something being a power of $5$. I have also tried to use the binomial expansion by noting that $5=2+3$, but I quickly ran out of ideas what to do with it. 
Any help would be appreciated! 

Let's try out some values of $a$.  The tl, dr:  we are forced $(a,b,c)=(1,1,1)$ when $a=1$ and to $(a,b,c)=(2,0,1)$ when $a=2$; then by using Pythagorean-triple-like solutions we fi n.v d these constraints eliminate all possibilities for larger $a$ except $(a,b,c)=(4,2,2)$.
$a=0$:
We can't have a power of $3$ and a power of $5$, both odd, differing by $1$.  No solution.
$a=1$: 
We have
$2+3^b=5^c$
For this to hold $\bmod 8$ we must have $b$ and $c$ odd.  With that constraint, if $b\ge 2$ the above equation requires $b\equiv c\equiv 1\bmod 6$ to hold $\bmod 7$, whereas $c\equiv 5\bmod 6$ would be needed $\bmod 9$.  These requirements are contradictory so $b\ge 2$ fails.  With $b$ odd we accept $(a,b,c)=(1,1,1)$ as the only solution with $a=1$.
$a=2$:
This gives
$4+3^b=5^c$
For this to hold $\bmod 8$ we need $b$ even and $c$ odd.  The latter causes failure $\bmod 3$ unless $b=0$, therefore this case admits only $(a,b,c)=(2,0,1)$.
$a\ge 3, \text{ odd}$
Now it gets meaty.  When $a\ge 3$ the power of $2$ becomes fixed $\bmod 8$ and the only way for the equation to hold now is to render $b$ and $c$ both even.
We then have a primitive triple of the form
$2p^2+q^2=r^2, 2p^2=(r+q)(r-q)$
where $p=2^{(a-1)/2}q=3^{b/2}, r=5^{c/2}$.  Then $r+q$ and $r-q$ must have a greatest common divisor of $2$, their product must be twice a square, and $q$ and $r$ must both be odd.  We are forced to this solution
$p=2mn, q=|m^2-2n^2|, r=m^2+2n^2$
Since $p$ must be a power of $2$, the primitive solution must have $m=1$ or $n=1$;  then we observe that $q$ (a power of $3$) must differ from $r$ (a power of $5$) by either $2$ or $4$.  From the earlier cases this forces $r=5$, and then we must have $2^a+3^b=r^2=25$ therefore no solution with $a$ odd ($2^a$ being twice a square).
$a\ge 4, \text{ even}$
This goes similarly to the case just rendered above, except the power of $2$ is now a square instead of twice a square.  We have the primitive Pythagorean triple
$p^2+q^2=r^2, p^2=(r+q)(r-q)$
with $p=2^{a/2}q=3^{b/2}, r=5^{c/2}$ and the solution
$p=2mn, q=|m^2-n^2|, r=m^2+n^2$
This time $q$ and $r$ must differ by $2$ forcing $q=3,r=5$, leading to $(p,q,r)=(4,3,5)$ and then $(a,b,c)=(4,2,2)$.
