In any case there are only finitely many rational solutions. If I interpret your question correctly, you are looking for rational points on the curve $C$ given by $$ (x-y)^4(x+y)=4xy $$ where I have put $x=a^2$ and $y=b^2$. Taking $t = x/y$, we easily find that $x$ satisfies $$ x^3 = \frac{4t}{(t-1)^4(t+1)}, $$ so that $C$ is birational to the superelliptic curve $C'$ given by $$ \eta^3 = 4 \xi (\xi-1)^2 (\xi + 1)^2. $$ We thus see that $C'$ is a triple cover of $\mathbb{P}^1$ which is totally ramified in $4$ points (including one above infinity), so by Riemann--Hurwitz $C'$ has genus $2$, so has finitely many rational points.

In any case there are only finitely many rational solutions. If I interpret your question correctly, you are looking for rational points on the curve $C$ given by $$ (x-y)^4(x+y)=4xy $$ where I have put $x=a^2$ and $y=b^2$. Taking $t = x/y$, we easily find that $y$ satisfies $$ y^3 = \frac{4t}{(t-1)^4(t+1)}, $$ so that $C$ is birational to the superelliptic curve $C'$ given by $$ \eta^3 = 4 \xi (\xi-1)^2 (\xi + 1)^2. $$ We thus see that $C'$ is a triple cover of $\mathbb{P}^1$ which is totally ramified in $4$ points (including one above infinity), so by Riemann--Hurwitz $C'$ has genus $2$, so has finitely many rational points.