The positive integer solutions for $2^a+3^b=5^c$

If $a=0$, then it is clear there are no solutions. If $b=0$, then we need $2^a + 1 = 5^c$. It is easy to show in this case $a=2,c=1$ is the only solution by showing that we need $2^{a-2}|c$. When $c=0$ there are obviously no solutions.

Suppose $a=1$. Then $2 + 3^b = 5^c$ only has the solution of $b=1,c=1$. To show this check modulo $275$ to deduce $c=1$ and thus $b=1$ is forced.

Now, suppose $a \ge 3$. Then remark that by checking modulo $4$ we need $b$ to be even so let $b = 2b'$. So let's solve $2^a + 3^{2b'} = 5^c$. Checking modulo $8$ we get $c$ is even so let $c = 2c'$. Then: $$2^a = (5^{c'} - 3^{b'})(5^{c'}+3^{b'})$$ We get $5^{c'} - 3^{b'} = 2^m, 5^{c'}+3^{b'} = 2^n$ for some $m,n$. But then $2 \cdot 5^{c'} = 2^m + 2^n$, forcing $m=1$. Thus we need $5^{c'} - 3^{b'} = 2$. But we already showed this only has the solution $b' = 1, c' = 1$. Thus it follows the only solution where $a \ge 2$ is with $a=4, b = 2, c= 2$.

Putting every together, we have proven the only solutions are $(2,0,1), (1,1,1), (4,2,2)$

EDIT: I realize I forgot to do the case of $a=2$. So we need to solve $4 + 3^{2b'} = 5^c$. Modulo $275$ happens to work again to force $c=1$ and thus we get no solutions when $b$ is nonzero.

Another solution is $2^4 + 3^2 = 5^2$. That's probably all.

case : $b>2$ and $c>2$

A calculus with computer, modulo $N=2372625=5^3\times 19 \times 37 \times 3^3$, in $H=\mathbb Z/N\mathbb Z $ give

$\text{card}(<2>)=900$, $\text{card}(3^3\times <3>)=900$, $\text{card}(5^3\times<5>)=36$.

And $(<2>+(3^3\times <3>)) \cap 5^3\times <5>=\emptyset$

with $1\in <g>$ the subset of $H$ generated by $g$ with times ($\times$)

So if $b>2$ and $c>2$, there are not solutions.

case : $b\leq 2$ and $c>2$

Here we choose $N=5^3\times (2^{25}-1)$

so $(<2>+\{1,3,9\}) \cap 5^3\times <5>=\emptyset$

So if $b\leq 2$ and $c>2$ no solution.

case : $c \leq 2$ three solutions

Then there are only 3 solutions, not any more.