There is no such number. Suppose $\alpha>1$ is a real number such that $\lfloor \alpha^n \rfloor$ is a square for all $n\in {\Bbb N}$. Put $\beta=\sqrt{\alpha}$.

Now for each $n$ we have
$$
m^2 + 1 > \alpha^n \ge m^2
$$
for some integer $m$, so that taking square-roots
$$
m + \frac{1}{2m} > \beta^n \ge m.
$$
In other words $\beta^n$ has exponentially small fractional part, indeed at most $1/(2m) \approx 1/(2\beta^n)$.

A theorem of Pisot now states that $\beta$ must be a Pisot--Vijayaraghavan (or PV) number (see for example the Wikipedia page on PV numbers). That is $\beta$ is an algebraic integer $>1$ such that all its Galois conjugates are $<1$ in absolute value. Suppose that $\beta$ has degree $k$ and that $\beta_1$, $\ldots$, $\beta_{k-1}$ are its Galois conjugates. Now for all $n$, we must have
$$
\beta^n + \beta_1^n + \ldots +\beta_{k-1}^{n} \in {\Bbb Z},
$$
and from our assumption it follows that for large $n$
$$
|\beta_1^n + \ldots +\beta_{k-1}^n |\le \frac{3}{4\beta^n}.
$$

Write each $\beta_j$ in polar coordinates as $r_j e^{2\pi i\theta_j}$. Note that
$$
\beta r_1 \cdots r_{k-1} \ge 1,
$$
since this is the absolute value of the norm of $\beta$. Moreover by Dirichlet's theorem we may find arbitrarily large $n$ with $\Vert n\theta_j \Vert \le 1/10^6$ for all $1\le j\le k-1$. Then, for such $n$,
\begin{align*}
Re(\beta_1^n + \ldots + \beta_{k-1}^n) &\ge 0.99 (r_1^n +\ldots +r_{k-1}^n) \ge 0.99 (k-1) (r_1\cdots r_{k-1})^{n/(k-1)} \\
&\ge 0.99 (k-1) \beta^{-n/(k-1)},
\end{align*}
by AM-GM. That gives a contradiction.

The argument also shows that if $\alpha^n$ is within a bounded distance of a square, then $\beta$ again is a PV number, and moreover its degree $k$ must be $2$ and its norm must be $1$ in size. In other words, we must have $\beta = (r+ \sqrt{r^2 \pm 4})/2$ for some natural number $r$. This is in keeping with Jeremy Rouse's comment above.