Solve $a^3-5a+7=3^b$ over the positive integer Solve $a^3-5a+7=3^b$ over the positive integer

I don't know how to solve such equation, please help me. Thanks

When $b=1$, we already know the result.
I can prove when $3\mid b$, there is no solution.
You may see 
if we try $a = 3^{b/3}$, then
$a^3-5a+7 = 3^b-5\cdot3^{b/3}+7<3^b$, if $b\ge3$.
If we try $a = 3^{b/3}+1$, then
$a^3-5a+7 = 3^b+3\cdot 3^{2b/3}-2\cdot 3^{b/3}+3>3^b$, if $b\ge 3.$
Thus $3^{b/3}<a<3^{b/3}+1$, there is no $a$ satisfying this, when $3\mid b$.
