Positive integers $a,b$ satisfying $a^3+a+1=3^b$

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).