$(am+b)^2+(a+bn)^2=a^2m^2+b^2+2amb+a^2+b^2n^2+2abn+a^2+b^2 \implies a^2m^2+2amb+b^2n^2+2abn$ needs to be an integer set $a=\frac{p}{q}$ , for $p,q\in \mathbb{Z}$ and $\gcd(p,q)=1$ similarly $b=\frac{r}{s}$ , for $r,s \in \mathbb{Z} , \gcd(r,s)=1$ so we must have $q^2\cdot s| s\cdot p^2\cdot m^2+2p\cdot r\cdot m \cdot q$ and $qs^2|r^2\cdot n^2 \cdot q+2p\cdot r\cdot s\cdot n \qquad \qquad (\spadesuit)$ if we take $m=p\cdot t$ and $n= q\cdot u$ s.t. $\gcd (t,q^2s)=1$ and $\gcd(u,qs^2)=1$ then clearly we can observe that $(\spadesuit)$ don't holds good hence we get not necessarily $(am+b)^2+(a+nb)^2$ is an integer for $a,b\in \mathbb{Q}$ and $m,n \in \mathbb{Z}$ $\blacksquare$

We claim the answer is NO, It is not possible to find such $m,n$ for; $a=\frac{p}{q}$, $b=\frac{r}{s}$ $q,s$ have prime factors of form $4k+3$, $q | r$ $(q,s)=1$ Note the given condition is equivalent to $a^2(m^2+1)+2ab(m+n)+b^2(n^2+1)$ is integer , so it suffices to prove $v_p(a^2(m^2+1)+2ab(m+n)+b^2(n^2+1)) <0$ for some prime $p$ Choose a prime factor $a$ of $q$ of form $4k+3$ then $v_p(a^2(m^2+1))<0$ while $v_p(b^2(n^2+1))>0$ and $v_p(2ab(m+n)) \geq 0$ Hence the claim $\blacksquare$