There are only a few cases to consider: \begin{tabular}{llllllllll} $837-n$ & 1 & 8 & 27 & 64 & 125 & 216 & 343 & 512 & 729 \\ $837+n$ & 1673 & 1666 & 1647 & 1610 & 1549 & 1458 & 1331 & 1162 & 945 \end{tabular}So $n=837-343=494$ is the only solution.

We can easily see that 837-n is a constantly decreasing term therefore we have to only see values for which it is positive and a cube

Write $837-n=x^3$ $837+m=y^3 \Rightarrow$ $y>x$ and $(x+y)(x^2-xy+y^2)=2 \cdot 3^3 \cdot 31$ Casework gives $(x, y) = (7, 11) \Rightarrow n=494$