Lemma 1. If $m$ is a positive integer and $\sqrt m$ is rational, then $\sqrt m$ is an integer.
Proof. Easy.
Lemma 2. If $m,n$ are positive integers and $\sqrt m+\sqrt n$ is rational, then both $\sqrt m$ and $\sqrt n$ are integers.
Proof. Say $\sqrt m+\sqrt n=x\in\Bbb Q$. Then
$$\sqrt m-\sqrt n=\frac{m-n}{x}$$
is rational and so is
$$\sqrt m=\frac{(\sqrt m+\sqrt n)+(\sqrt m-\sqrt n)}{2}\ ,$$
and likewise $\sqrt n\,$. By lemma 1, $\sqrt m$ and $\sqrt n$ are integers.
Now suppose that
$$\sqrt a+\sqrt b+\sqrt c=\sqrt s\ ,$$
where $a,b,c,s$ are positive integers and $s$ is squarefree. Squaring and rearranging,
$$2\bigl(\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\bigl)=s-a-b-c\ .$$
Now add to this equation the identity $2\sqrt a\sqrt a=2a$ and factorise to obtain
$$2\sqrt{bc}+2\sqrt{as}=s+a-b-c\ .$$
By lemma 2, we see that $\sqrt{as}$ is an integer; since $s$ is squarefree, $a$ must be a square times $s$, say $a=p^2s$. Similarly $b=q^2s$ and $c=r^2s$, so
$$p\sqrt s+q\sqrt s+r\sqrt s=\sqrt s\ ,$$
but as $p+q+r>1$, this is impossible.
Supposing $m$ is not a perfect square, then $m=n^2+k$, where $n^2$ is the largest perfect square less than $m$. Without loss of generality, if $k>n$ we can take $m_0=m-n$ and $k_0=k-n$, otherwise $m_0=m, k_0=k$.
Then we can see that $f^2(m_0) = n^2+k_0+2n = (n+1)^2+(k_0-1)$.
Taking $m_1=f^2(m_0)$ and $k_1=(k_0-1)$ we can see the same process applies relative to $(n+1)^2$ and so in a total of $2k_0$ applications of $f$ we will have a perfect square, $f^{2k_0}(m_0) = (n+k_0)^2$.
Additional observation: Note that once a square is found, $s_0^2 = f^d(m)$, the same process can be applied to $f^{d+1}(m) = s_0^2+s_0$, which will then give another perfect square at $f^{d+1+2s_0}(m) = (2s_0)^2$.
Thus there are an infinite number of perfect squares in the given sequence, of the form $(2^as_0)^2$, where $a$ is a non-negative integer. This also means there is at most one odd square in the sequence, which only occurs if $m_0$ is odd (or if $m$ itself is an odd square).
Best Answer
A square is equal to $0$ or $1$ mod $3$ but not equal to $2$ mod $3$. Can you take it from here?