Your curve can be written as $Y^2=X^4+X^3-2,$ where $Y=4n$ and $X=q^2.$ This Diophantine equation satisfies Runge's condition, so this is relatively easy to handle and one obtains that there are only finitely many integral solutions (see Poulakis-Quartic). You may also consider it as a genus 1 curve and there are techniques to determine all integral points on such curves (see Tzanakis-Quartic).
To make it a bit more explicit, let $P(X)=X^4+X^3-2, P_1(X)=4X^2+2X$ and $P_2(X)=4X^2+2X-1.$ Here we get that $16P(X)-P_1(X)^2=-4X^2-32$ and $16P(X)-P_2(X)^2=4X^2 + 4X - 33.$ Hence $$(4X^2+2X-1)^2<16P(X)=(4Y)^2<(4X^2+2X)^2$$ if $X\notin [-4..3].$ That is we have a contradiction, since $(4Y)^2$ is supposed to be between two consecutive squares. It remains to deal with the values $X\in [-4..3].$ The only solution is given by $$(X,Y)=(1,0).$$ Thus $n=0$ and $q=\pm 1$ (you look for positive solutions only, so $q=1$ remains). That was the Runge approach.
The elliptic curve part can be done by the program package Magma (see Magma), you simply type
IntegralQuarticPoints([1,1,0,0,-2],[1,0]);
and you get
[
[ 1, 0 ]
].
Here $[1,1,0,0,-2]$ comes from the degree 4 polynomial, these are the coefficients and $[1,0]$ is a point on the curve.