Positive integers $a,b$ satisfying $a^3+a+1=3^b$ How to prove that $a=b=1$ is the only positive integer solution to the following Diophantine equation?$$a^3+a+1=3^b$$  

Partial solution, Check the equation mod 9 $\big(\forall b>1,3^b\equiv 0 \pmod 9\big)$. Nine cases to check:
$a=0 \pmod 9, a^3+a+1\equiv 1 \pmod 9\\ 
a=1 \pmod 9, a^3+a+1\equiv 3 \pmod 9 \\
a=2 \pmod 9, a^3+a+1\equiv 2 \pmod 9\\
a=3 \pmod 9, a^3+a+1\equiv 4 \pmod 9\\
a=4 \pmod 9, a^3+a+1\equiv 6 \pmod 9\\
a=5 \pmod 9, a^3+a+1\equiv 5 \pmod 9\\
a=6 \pmod 9, a^3+a+1\equiv 7 \pmod 9\\
a=7 \pmod 9, a^3+a+1\equiv 0 \pmod 9\\
a=8 \pmod 9, a^3+a+1\equiv 8 \pmod 9 $
so the only cases possible greater than a=b=1, is if a, is 7 mod 9, and b>1 . plugging that in we get:
$(9j+7)^3+(9j+7)+1 = 729j^3 + 1701j^2 + 1332j + 351$ ( done with PARI/GP)  which reduces to 26j when done mod 27. Without j a multiple of 27, this won't be 0 mod 27 which $3^b$ is for b>2.
