Here's a reformulation of the conjecture, and a heuristic which suggests that the conjecture is false.  Note that $a+b$ must be odd.  Put $\ell = a+b$  (and $\ell$ is odd) and $m=a-b$ which is also odd.  The condition that $b<a<2b$ now translates to $0<m < \ell/3$.  The divisibility condition of the question translates nicely to the condition $\ell^2$ divides $m^3-2$.  

So the question really is can $m^3-2$ have a square factor of size at least $(3m)^2$?   Here's a heuristic that indicates why this should be possible.  Given $\ell$ odd, the number of solutions $\pmod {\ell^2}$ to $x^3-2\equiv 0 \pmod{\ell^2}$ is given by a multiplicative function $f(\ell)$.  Here if $p\ge 5$ then $f(p^k)$ is simply the number of solutions to $x^3\equiv 2 \pmod p$ which is $0$, $1$ or $3$ (with asymptotic densities as in Chebotarev).  For $p=3$ one should work a tiny bit harder.  Then the question is whether one of these $f(\ell)$ solutions $\pmod{\ell^2}$ happens to lie below $\ell/3$.  We may expect that probability to be about $f(\ell)/(3\ell)$.  Thus the expected number of $\ell$ below $x$ for which $\ell^2$ divides $m^3-2$ (with $m< \ell/3$) should be about 
$$ 
\sum_{\ell \le x; \ell \text{ odd}} f(\ell)/(3\ell) \sim C \log x,
$$ 
for some positive constant $C$ -- with a little calculation, using the Dedekind zeta function one should be able to calculate $C$.   This suggests that there are infinitely many counterexamples.  

I didn't find any with $m$ below $10000$ (I searched for large square factors of $m^3-2$), but my guess would be that there is a reasonable chance of finding counterexamples with $m$ in the millions, and a very good chance if $m$ is in the billions.