First of all, it's easy to see that $\lambda$ must be nonnegative, since otherwise $\left \lfloor \lambda^{4n-1} \right \rfloor$ is negative and so clearly isn't a perfect square. Therefore, we can assume WLOG that $\lambda \ge 0.$ If $\lambda < 2$, then $\left \lfloor \lambda \right \rfloor$ is either $0$ or $1$, and so we're done. Else, let's assume that $\lambda \ge 2.$ Call a positive real number $x$ $\mathbf{good}$ if $\left \lfloor x \right \rfloor$ is a perfect square (including $0$). The following lemma will prove useful in solving the problem: $\mathbf{Lemma.}$ For any real number $r \ge 2$, if $r^2, r^3, r^4$ are all good, then so is $r$. $\mathbf{Proof.}$ Let $\left \lfloor r^2 \right \rfloor = x^2,$ where $x \in \mathbb{N}.$ Then, we have that $x^4 \le r^4 < (x^2 + 1)^2,$ and so since $r^4$ is good we have that $\left \lfloor r^4 \right \rfloor = x^4.$ Now, since $r \ge x$, we have that $r^3 \ge r^2 \cdot x \ge x^3.$ Furthermore, we know that since $r^4 < x^4 + 1,$ we have that $r^3 < \frac{x^4 + 1}{r} < \frac{x^4 + 1}{x} < x^3 + \frac1x < x^3 + 1,$ giving us that $\left \lfloor r^3 \right \rfloor = x^3$ is a perfect square. This clearly implies that $x$ is a perfect square, and so since we clearly have $\left \lfloor r \right \rfloor = x$, the lemma is proven. $\blacksquare$ By the lemma, we can easily show by downwards induction that $\lambda^t, \lambda^{t+1}, \cdots, \lambda^{4n}$ are all good, for any $1 \le t \le n.$ For $t = n$, we can apply the lemma for $r = \lambda^n.$ For the inductive step, if $\lambda^{t+1}, \cdots, \lambda^{4n}$ are all good, then applying the lemma on $r = \lambda^t$ gives that $\lambda^t$ is too. Eventually, we get that $\lambda$ is good, as desired. $\square$

Here is an another way to apply induction. We can assume $\lambda < 2$ as in above solution. Let $a^2\leq \lambda^{2n+2} < a^2+1$ and $b^2\leq \lambda^{4n} <b^2+1$. With this we have, $$ a\leq \lambda^{n+1} <\sqrt{a^2+1} <a+1 \implies \lfloor \lambda^{n+1}\rfloor =a=c^2 $$$$ b\leq \lambda^{2n} <\sqrt{b^2+1} <b+1 \implies \lfloor \lambda^{2n}\rfloor =b=d^2 $$Thus we get $$ c^2\leq \lambda^{n+1}< \sqrt{c^4+1} $$$$ d\leq \lambda^n < (d^4+1)^{\frac{1}{4}} $$Hence, $$ c^2\cdot d\leq \lambda^{2n+1}<(c^4+1)^{\frac{1}{2}} \cdot (d^4+1)^{\frac{1}{4}}\leq \frac{1}{2^{\frac{1}{4}}} \cdot (c^2d+1)\implies \lfloor \lambda^{2n+1}\rfloor=c^2d $$So $\lfloor \lambda^{n}\rfloor=d$ must be a perfect square. We can get $n$ from $(4n,2n+2,2n+1,2n,n+1)$ so $\lfloor \lambda\rfloor$ must be a perfect square too by induction.

beautiful way!

We induct on $n$ with the base case being $n = 1$. For $n = 1$, let \[\lfloor \lambda^2 \rfloor = m^2.\] Then $m^4 \leq \lambda^4 < (m^2+1)^2$ giving us $\lfloor \lambda^4 \rfloor = m^4.$ Also as $m^2 \leq \lambda^2 <(m+1)^2$ this gives $\lfloor \lambda \rfloor = m.$ Therefore, $$ \lambda^3 = \frac{\lambda^4}{\lambda} < \frac{m^4 + 1}{m} = m^3 + \frac1m$$ giving us $\lfloor \lambda^3 \rfloor = m^3.$ This implies $m$ is a perfect square. Hence $\lfloor \lambda \rfloor$ is a perfect square. Now it suffices to show that $\lfloor \lambda^n \rfloor$ is a perfect square as we can induct down. Let $\lfloor \lambda^{2n} \rfloor = m^2.$ This gives $\lfloor \lambda^{4n} \rfloor = m^4$ and $\lfloor \lambda^n \rfloor = m$. But now $\lfloor \lambda^{3n} \rfloor = m^3,$ giving us $m$ is a perfect square. So, $\lfloor \lambda^n \rfloor$ is a perfect square as needed.