Denote $n=m+p$ where $p$ is a positive integer. Claim: We can assume that $m,n$ are odd. Proof: Suppose that $m,n$ are even. We have $2v_2(m)=v_2(r)=2v_2(n)$. Let $m=2^xk$ and $n=2^xl$. Also $r=2^{2x}s$ The equations become $2^{2x}k^2+2^{2x}s=2^b$ and $2^{2x}l^2+2^{2x}s=2^a$ iff $k^2+s=2^b$ and $l^2+s=2^a$. If the inequality $ls>2k^2$ holds, then $2^{3x}ls>2^{2x+1}k^2$ also holds. Thus, proving the problem for odd $m,n$ is enough. \[m^2+r=2^a \ \text{and} \ m^2+r+2mp+p^2=2^b\]$p(2m+p)=2mp+p^2=2^a(2^{b-a}-1)$. If $a=1,$ then we have no solution. If $a=2,$ then $m=1$ and $r=3$ which yields no solution for $n^2+3=2^b$. Hence $a\geq 3$. If $v_2(p)\geq 2,$ then $v_2(2m+p)=1$ since $m$ is odd. $a=v_2(p)+1$ thus, $2^{a-1}|p$. $\frac{m^2+r}{2}|p\implies p\geq \frac{m^2+r}{2}$ Also by looking at $(mod \ 8),$ we get that $r\geq 7$. Hence \[nr=(m+p)r\geq 7(m+\frac{m^2+r}{2})>\frac{7}{2}m^2>2m^2\]If $v_2(p)=1,$ then let's write $p=2t$ where $t$ is an odd integer. Note that $r\equiv 7(mod \ 8)$ $m^2+r|4mt+4t^2\iff m^2+r|4m+4t\implies 4t\geq m^2-4m+r\geq m^2-4m+7$ By multiplying both sides by $\frac{7}{2}$ we get that $14t\geq \frac{7}{2}(m^2-4m+7)=\frac{7}{2}m^2-14m+\frac{49}{2}>2m^2-7m$ hence $nr\geq 14t+7m>2m^2$ as desired.$\blacksquare$