We first prove the statement for $n = 1$. Write $x = a + r$, with $a \ge 1$ an integer and $0 \le r < 1$. Suppose $\lfloor x^2 \rfloor, \lfloor x^3 \rfloor$ and $\lfloor x^4 \rfloor$ are squares. Then we have $a \le x < a + 1$, from which it follows that $a^2 \le x^2 < (a+1)^2$. Hence, $x^2$ is squeezed between two consecutive squares. However, $\lfloor x^2 \rfloor$ is a square, hence the only possibility is that $\lfloor x^2 \rfloor = a^2$. We conclude that $a^2 \le x^2 < a^2 + 1$.

Completely analogously, we also get that $(a^2)^2 \le x^4 < (a^2 + 1)^2$ and hence $\lfloor x^4 \rfloor = a^4$. We conclude that $a^4 \le x^4 < a^4 + 1$.

Moreover, we have that $x^3 \ge a^3$. Now suppose that $x^3 \ge a^3 + 1$, i.e.
$$x^4 \ge x(a^3 + 1) = (a + r)(a^3 + 1) = a^4 + r a^3 + a + r \ge a^4 + a \ge a^4 + 1,$$
which gives a contradiction. Hence, $x^3 < a^3 + 1$, which yields that $\lfloor x^3 \rfloor = a^3$. This is also a square, hence $a$ must be a square itself. We see that $\lfloor x \rfloor$ is a square.

Now we will finish the proof with induction to $n$. The induction basis has just been proved. Now let $k \ge 1$ and suppose that the statement is proved for $n = k$. Consider a real number $x \ge 1$ with the property that $\lfloor x^{k+2} \rfloor, \lfloor x^{k+3} \rfloor, \dots, \lfloor x^{4k+4} \rfloor$ are all squares. In particular, $\lfloor x^{2(k+1)} \rfloor, \lfloor x^{3(k+1)} \rfloor$, and $\lfloor x^{4(k+1)} \rfloor$ are all squares. We can now apply the case $n = 1$ on $x^{k+1}$ (which is a real number greater than or equal to 1) and find that $\lfloor x^{k+1} \rfloor$ is also a square. Now we know that $\lfloor x^{k+1} \rfloor, \lfloor x^{k+2} \rfloor, \dots, \lfloor x^{4k} \rfloor$ are all squares and using the induction hypothesis, we obtain that $\lfloor x \rfloor$ is a square as well. This completes the proof by induction. $\square$