Integer solutions of $a^6 + 4 b^3 = 1$ This question is simplified case of 
Integer solutions of $a^6 + 4 b^3 = c^6$.
Is there integer solutions for
$$a^6 + 4 b^3 = 1$$
The way I am trying to prove is based on Fermat's Theorem, but I can not get the final result.

From $a^6+4b^3=1$, we see that $a$ must be odd. Let's rewrite the equation as 
$$(a^3-1)(a^3+1)=-4b^3$$ 
Since $a^3$ is odd, we have $\gcd(a^3-1,a^3+1)=2$, so we have
$$\begin{align}
a^3-1&=2c^3\\
a^3+1&=2d^3
\end{align}$$
where $cd=-b$ with $\gcd(c,d)=1$. But this implies
$$a^3=c^3+d^3$$
which, as Fermat observed, has no nontrivial integer solutions. So we're left with $(a,b)=(\pm1,0)$ as the only integer solution.
Remark (added later): As the OP brilliantly observes, there is really no need to rely on Fermat.  If you subtract the two equations instead of adding them, you get $d^3-c^3=1$, which immediately eliminates anything nontrivial, since cubes have a hard time differing by $1$. 
