Lemma. *There are infinitely many positive integers $n$ such that*
$$f(n) \geq n^{\frac{13}{10}}.$$
Proof. Letting $(m,n) = (m,f(m))$ to obtain $f(f(m)) + 1400m^2 \mid f(m)^2 + f(f(m))$ yielding $f(m) > m$. Plugging $n=1$ to obtain $f(1) + 1400m^2 \mid 1 + f(f(m))$ yielding $f(f(m)) > m^2$. Assume to the contrary that for all but finitely many $n$ we have $f(n) \le n^{\sqrt{2}}$. Since $f(n) > n$ we would obtain $n^2 < f(f(n)) \le f(n)^{\sqrt{2}}$ yielding $f(n) > n^{\sqrt{2}}$, a contradiction. Hence, $f(n) > n^{\sqrt{2}} > n^{\frac{13}{10}}$ for infinitely many $n$. This completes our proof.

Let us denote the set of such $n$ by $T$. Letting $t \in T$ and $a$ be a fixed positive integer. Plugging $(m,n) = (t,1)$, $(t,a)$ to obtain $t^2 + A = C(f(t) + 1400)$ and $t^2 + B = D(f(t) + 1400a^2)$ for some positive integers $C, D$ whilst $A = f(f(1))$, $B = f(f(a))$. Choose $t$ such that $A, B < t^{\frac{7}{10}}$ it follows that
$$1400CD(1 - a^2) - (AD - BC) = t^2(C - D).$$
The left side is $O(t^{\frac{7}{10}})$ while the right side is $O(t^2)$ unless $C = D$. Hence, for all large enough $t$, $C = D$ and therefore, $C = \frac{A-B}{1400(1-a^2)}$, $f(t) = \frac{t^2+A-1400C}{C}$. Notice that $C$ is a function of $t$ hence, $C$ doesn't depend on $a$. Thus changing $a$ would not change $C$. Hence, for all positive integers $a, b$ :
$$\frac{f(f(a)) - f(f(1))}{1400(1 - a^2)} = \frac{f(f(b)) - f(f(1))}{1400(1 - b^2)}.$$
Implying that $f(f(a)) = ra^2 + s$, for some constants $r, s$. That is,
$$f(n) + 1400m^2 \mid n^2 + rm^2 + s,$$
Yielding
$$f(n) + 1400m^2 \mid 1400n^2 - rf(n) + 1400s.$$
Choose $m$ large enough to obtain that $1400n^2 - rf(n) + 1400s = 0$. That is,
$$f(n) = \frac{1400n^2 + 1400s}{r},$$
But then $f(f(n))$ must be a polynomial of degree 4 in $n$. This contradicts to what we've already obtained. Hence, there is no such a function. ■