Let us analyze the sequence $a_i$ defined by $a_1 = m$ and $a_{i+1} = a_i + \lfloor \sqrt{a_i} \rfloor$.

Suppose at some step $a_k$ is a perfect square, say $a_k = n^2$ for some integer $n \geq 1$. Then:
$$a_{k+1} = a_k + \lfloor \sqrt{a_k} \rfloor = n^2 + \lfloor n \rfloor = n^2 + n = n(n+1)$$

Now, consider the next terms:
$$a_{k+2} = a_{k+1} + \lfloor \sqrt{a_{k+1}} \rfloor$$
But $a_{k+1} = n(n+1)$, and $\sqrt{n(n+1)}$ is between $n$ and $n+1$.
Specifically, $n < \sqrt{n(n+1)} < n+1$ for $n \geq 1$, so $\lfloor \sqrt{n(n+1)} \rfloor = n$.
Therefore:
$$a_{k+2} = n(n+1) + n = n^2 + 2n = (n+1)^2 - 1$$

Now, $a_{k+3} = a_{k+2} + \lfloor \sqrt{a_{k+2}} \rfloor$
But $a_{k+2} = (n+1)^2 - 1$, so $\sqrt{a_{k+2}} = \sqrt{(n+1)^2 - 1}$.
Since $(n+1)^2 - 1 < (n+1)^2$, $\sqrt{a_{k+2}} < n+1$, and $\sqrt{a_{k+2}} > n$ (since $(n+1)^2 - 1 > n^2$ for $n \geq 1$), so $\lfloor \sqrt{a_{k+2}} \rfloor = n$.
Thus:
$$a_{k+3} = (n+1)^2 - 1 + n = n^2 + 2n + n - 1 = n^2 + 3n - 1$$

But let's look for the next perfect square. Let us try to see if the sequence can reach $(n+1)^2$.
From above, $a_{k+2} = (n+1)^2 - 1$, $a_{k+3} = n^2 + 3n - 1$.

Let us try to generalize. For large $a_i$, $\lfloor \sqrt{a_i} \rfloor$ increases slowly compared to $a_i$, so the sequence grows slowly.

But more importantly, we can show that for any starting $m$, the sequence will eventually reach a perfect square, and then the above process repeats.

Let us prove that for any $a_i$, there exists $j > i$ such that $a_j$ is a perfect square.

Let $a_i$ be arbitrary. Let $n = \lfloor \sqrt{a_i} \rfloor$. Then $n^2 \leq a_i < (n+1)^2$.

If $a_i = n^2$, we are done. If not, $a_i = n^2 + r$ for $1 \leq r < 2n + 1$.

At each step, $a_{i+1} = a_i + \lfloor \sqrt{a_i} \rfloor$. Since $a_i \geq n^2$, $\sqrt{a_i} \geq n$, so $\lfloor \sqrt{a_i} \rfloor \geq n$.

Let us consider the sequence of $a_i$ modulo $n$:

Let us try to reach $(n+1)^2 = n^2 + 2n + 1$.

From $a_i = n^2 + r$, after $s$ steps, $a_{i+s} = n^2 + r + s n$ (since $\lfloor \sqrt{a_{i+t}} \rfloor = n$ for $a_{i+t} < (n+1)^2$).

We want $a_{i+s} = (n+1)^2 = n^2 + 2n + 1$.
So:
$$n^2 + r + s n = n^2 + 2n + 1$$
$$r + s n = 2n + 1$$
$$s n = 2n + 1 - r$$
$$s = \frac{2n + 1 - r}{n}$$

Since $1 \leq r < 2n + 1$, $2n + 1 - r$ is positive and less than $2n$.
So $s$ is an integer if and only if $n$ divides $2n + 1 - r$.

But even if $s$ is not integer, after $t$ steps, $a_{i+t}$ will reach or exceed $(n+1)^2$, at which point $\lfloor \sqrt{a_{i+t}} \rfloor = n+1$.

Therefore, after a bounded number of steps, the sequence will cross from $n^2$ to $(n+1)^2$.

Now, consider the difference $a_{i+t} - (n+1)^2$. For some $t$, $a_{i+t}$ will be exactly $(n+1)^2$ if $a_i$ is congruent to $n^2$ modulo $n$.

But even if not, after passing $(n+1)^2$, the process repeats with $n$ replaced by $n+1$.

Therefore, for any starting $m$, the sequence will eventually hit a perfect square, and then the process repeats. Thus, the sequence contains infinitely many perfect squares.

Therefore, the sequence $a_i$ contains infinitely many perfect squares.