Assume there's a non-integral $x$ which satisfies the stated criteria. For $n = 1$, for some integer $m \ge 1$, we'd have
$$\lfloor x\rfloor = m^2 \;\;\to\;\; x = m^2 + r, \;\; 0 \lt r \lt 1 \tag{1}\label{eq1A}$$
With $n = 2$, we get
$$(m^2)^2 \lt x^2 = (m^2 + r)^2 \lt (m^2 + 1)^2 \tag{2}\label{eq2A}$$
Thus, for $\lfloor x^2\rfloor$ to be a perfect square requires that
$$\lfloor x^2\rfloor = (m^2)^2 \;\;\to\;\; (m^2+r)^2 \lt (m^2)^2 + 1 \tag{3}\label{eq3A}$$
Have $n$ double each time, so $n = 2^k$. Using the first $2$ terms of the binomial theorem expansion, and with large enough $k$, we have
$$\begin{equation}\begin{aligned}
& (m^2+r)^{2^{k}} - (m^{2^k}+1)^2 \\
& \gt (m^2)^{2^k} + 2^{k}(m^2)^{2^k-1}r - (m^{2^k})^2 - 2(m^{2^k}) - 1 \\
& = m^{2^{k+1}} + 2^{k}rm^{2^{k+1}-2} - m^{2^{k+1}} - 2m^{2^k} - 1 \\
& = 2m^{2^k}\left(2^{k-1}rm^{2^{k}-2}-1\right) - 1 \\
& \gt 0
\end{aligned}\end{equation}\tag{4}\label{eq4A}$$
This shows $\left\lfloor x^{2^k}\right\rfloor \ge (m^{2^k}+1)^2$ for all sufficiently large $k$. Note that, for any $k$, we have $x^{2^k}\gt \left(m^{2^{k}}\right)^2$, and $\left\lfloor x^{2^k}\right\rfloor$ can't be any value between $\left(m^{2^k}\right)^{2}+1$ and $\left(m^{2^k}+1\right)^{2}-1$, inclusive. Thus, for some $k$, there must be a switch between the floor value being $(m^{2^k})^2$, and then for $k+1$ it being at least $(m^{2^{k+1}}+1)^2$, i.e., that
$$(m^2+r)^{2^{k}} \lt (m^{2^k})^2+1, \;\; (m^2+r)^{2^{k+1}} \ge \left((m^{2^k})^2+1\right)^2 \tag{5}\label{eq5A}$$
However, squaring the left side above gives $(m^2+r)^{2^{k+1}} \lt \left((m^{2^k})^2+1\right)^2$, contradicting the right side above. This shows the original assumption that there's such a non-integral $x$ must be false.
Best Answer
For all naturals $n$ we have $$f(n+1)-f(n)=44\cdot10^{2n}+8\cdot{10^n}.$$
Hence, $$\frac{f(n+1)-f(n)}{16}=11\cdot25\cdot10^{2(n-1)}+5\cdot10^{n-1}\in\mathbb N.$$
It follows that $$f(n)\equiv f(1)\equiv8\pmod{16}$$ for all $n\in\mathbb N$.
Because every square is equivalent to $0,1,4$ or $9$ modulo $16$, it follows that $f(n)$ is never a square.