I claim the only answer are $(a,b)=(2^i,2^i)$, where $i\ge 0$ is arbitrary. If $d={\rm gcd}(a,b)$ then we get $d=p^i$ for some prime $p$ and non-negative integer $i$, so we assume $(a,b)=1$. Further, let $u=a^2$ and $v=b^2$. We are left with $(u,v)=1$ and for some prime $p$ and $k\in\mathbb{Z}_+$, \[ p^k = f(u,v)f(v,u),\quad\text{where}\quad f(u,v)=u^3+21u^2v+35uv^2+7v^3. \]I now show $(u,v)=(1,1)$ (with $p=2$) is the only solution. To that end, we verify $p\mid f(u,v)+f(v,u)=8(u+v)(u^2+6uv+v^2)$. If $p\mid u+v$ then $f(u,v)\equiv 8u^3\pmod{p}$. As $p\nmid u$ (otherwise $p\mid u+v$ gives $p\mid v$, which contradicts with $(u,v)=1$), we have $p\mid 8$. Likewise if $p\mid u^2+6uv+v^2$ then $p\mid 7u^3+42u^2v+7uv^2$, so $p\mid 7u^2v -14uv^2-v^3 = v(7u^2-14uv-v^2)$. Once again, $p\nmid v$ (otherwise $p\mid u$), so $p\mid 7u^2-14uv-v^2$. Now, as $p\mid u^2+6uv+v^2$, we get $p\mid 8u^2-8uv=8u(u-v)$. So if $p\nmid 8$ then $p\mid u$ (impossible) or $p\mid u-v$, which yields $u^2+6uv+v^2\equiv 8v^2\pmod{p}$. In any event, we get that $p=2$. Now, if $u=v$ then $u=v=1$ is the only solution which indeed works. So, assume $u\ne v$ and let $u>v$ without loss. As $f(v,u)-f(u,v)>0$, we get $f(u,v)\mid f(v,u)$. Further, as $f(v,u)<7f(u,v)$ and both are powers of two, we have either $f(v,u)=2f(u,v)$ or $f(v,u)=4f(u,v)$. In the former case, we get $5u^3 = 7u^2v+49uv^2+13v^3$. Note that $7\nmid v$ as otherwise $7\mid u$. Inspecting both sides modulo $7$, we find $(u/v)^3\equiv -3\pmod{7}$, which is clearly impossible. The case $f(v,u)=4f(u,v)$ is handled analogously. From here, we get \[ (a,b) = (2^i,2^i) \]where $i\ge 0$ is arbitrary, is indeed the only solution.