Assume $d\neq a,b$. One of $\gcd\{ d,a \} $ and $\gcd\{ d,b \}$ must be $\geq \sqrt{d}>m$, but both $\lvert d-a\rvert<m$ and $\lvert d-a\rvert<m$ so this can't happen; the only solutions are $d=a$ and $d=b$.

Clearly ,$d=a,b$ works, we will consider $d\neq a,b$ Since $d\mid ab$ there exist $k\in \mathbb{Z}^+$ that $ab=dk$ It is well known that there exist $p,q,r,s$ that $a=pq,b=rs,d=pr,k=qs$ where $gcd(q,s)=1$ So we get $m^2<pq,pr,rs<m^2+m$ and $r\neq q,p\neq s$ We get $|p||q-r|<m$ give us $p<m$ and $|r||p-s|<m$ give us $r<m$, so $pr<m^2$ contradiction So all possible value of $d$ are $a$ and $b$

Another solution. Let $a<b$ w.l.o.g., and set $a=m^2+u$ and $b=m^2+v$ with $1\le u<v\le m-1$. Assume $d=m^2+\ell$. Note that $m^2\equiv -\ell\pmod{d}$, hence $a =m^2 + u \equiv u-\ell \pmod{d}$ and $b\equiv v-\ell\pmod{d}$. This forces $d\mid |(u-\ell)(v-\ell)|$. If $u=\ell$ or $v=\ell$, we obtain $d=a,b$. If not, then $1\le |u-\ell|,|v-\ell|\le m-2$, forcing $d=m^2+\ell \le (m-2)^2$. This is a contradiction since $\ell>0$.