Integral solutions to $56u^2 + 12 u + 1 = w^3$

Multiplying by $14^3$, we obtain $$14^4\cdot 2^2 u^2 + 14^3 \cdot 12u + 14^3 = (14w)^3 \implies (14(28u+3))^2+980 = (14w)^3$$ This is a Mordell equation of the form $Y^2 = X^3-980$, which has $5$ solutions given by $(14,42)$, $(21,91)$, $(29,153)$, $(126,1414)$ and $(326,5886)$.

Of this only $(14,42)$ has $Y$ of the form $14(28u+3)$. Hence, the only solution is $u=0$ and $w=1$.