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?$$a^3+a+1=3^b$$  
 A: Yes, $a=b=1$ is the only solution.
As Mike Bennett suggests in the comments,
this can be proved by asking more generally for
integer points on the elliptic curves $y^2 = x^3 + x + 1$ [for $b$ even,
with $(x,y) = (a,3^{b/2})$] and $3y^2 = x^3+x+1$ [for $b$ odd, with
$(a,y) = (a,3^{(b-1)/2})$].
Siegel's theorem shows that there are
only finitely many solutions; his proof is ineffective, but
since then new techniques have been developed and implemented to 
provably list all solutions.
As Gerry Myerson suggests in his comment, these curves
are simple enough that the computation of their integer points
has already been done.  These days the most extensive online databases
for such results are collected in the LMFDB,
so I looked there.
The first curve, $y^2 = x^3 + x + 1$, is 
496.a1;
there are two pairs of integral points, $(0,\pm 1)$ and $(72,\pm 611)$.
Only the first makes $y$ a power of $3$, and gives the solution $(a,b)=(0,0)$,
which is integral, but not positive as Ghartal asks.
The second curve, $3y^2 = x^3 + x + 1$, is not in Weiestrass form,
but we can take $X=3x$ and find that 
$$
X^3+9X+27 = 27(x^3+x+1) = 81y^2 = (9y)^2,
$$
so $(3x,9y)$ is an integral point on the curve $Y^2 = X^3 + 9X + 27$.
This is curve
2232.j1
in the LMFDB; its only integral points are $(X,Y) = (-2,\pm1)$ and $(3,\pm9)$.
The former yields $(a,b) = (-2/3,-3)$, a rational solution but again
not satisfying Ghartal's condition that $a,b$ be positive integers.
The latter yields the known solution $(a,b)=(1,1)$.  
This completes the determination of solutions of $a^3+a+1 = 3^b$
in positive integers.
It may be quite hard to give an elementary proof; e.g. once we have the
nontrivial solution with $b=1$ it's impossible to exclude further 
solutions by congruence conditions alone (congruence modulo
powers of $3$ only impose conditions on $a$ modulo powers of $3$,
which cannot combine with congruences to moduli coprime with $3$
to give a contradiction).
A: 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.
