This one is tough. I claim there are no solutions. To that end, set $d={\rm gcd}(m,n)$; set $m=dm_1$ and $n=dn_1$. Then $d^4m_1n_1(m_1^2+6m_1n_1+n_1^2)$ is a square. As $(m_1n_1,m_1^2+6m_1n_1+n_1^2)=1$, it follows that for some $a,b$, we get $m_1=a^2$ and $n_1=b^2$. Hence, it boils down proving that \[ a^4+6a^2b^2+b^4 \]is never a square if $(a,b)=1$. To that end, assume the contrary. Then $(a^2+b^2)^2+(2ab)^2 = c^2$ for some $a,b,c$. We now recall the characterization of Pythagorean triples, where the are two cases to consider. Case 1. For some $(m,n)=1$, $a^2+b^2=2mn$ and $2ab=m^2-n^2$. The second equation tells $m\equiv n\pmod{2}$, hence $4\mid 2ab$. Thus either $2\nmid a$ or $2\nmid b$. If this is the case, then the first equation tells $2\mid (a,b)$, a contradiction. Case 2. For some $(m,n)=1$, $a^2+b^2=m^2-n^2$ and $ab=mn$. Note that if $p\mid (m,n)$ is a prime then $p\mid a,b$, a contradiction. Hence $(a,b)=(m,n)=1$. From here, it follows that there exists $(u,v)=1$ such that $a=uv$, $u\mid m$ and $v\mid n$. This yields $m=ku$ and $n=\ell v$ form some $k,\ell$ with $(k,\ell)=1$. In particular, $b=k\ell$, and inserting these back we find that \[ k^2(u^2-\ell^2) = v^2(u^2+\ell^2). \]Multiply both sides by $u^2-\ell^2$ to get that $u^4-\ell^4$ is a square. Moreover as $(a,b)=1$, we get that $(u,\ell)=1$. This, however, is very well-known to be impossible, see e.g. last page in here. (I am sure there are other links here on AoPS.)

Just simplify grupyorum proof $(a^2+b^2)^2+(2ab)^2=c^2$ For $a\ne b$, multiply both sides by $(a^2+b^2)^2-(2ab)^2$ we get $(a^2+b^2)^4-(2ab)^4=[c(a^2-b^2)]^2$