$\sqrt{2^2} = 2$, $\sqrt{2^4} = 2^2 = 4$, $\sqrt{2^6} = 8 \neq 6$, $\ldots$ In general, if $t \in \mathbb{N}$, when is it true that
$$\sqrt{2^{2t}} = 2^t = 2t?$$

I know of course that it is true when $t - 1 = \log_{2}(t)$.  Are there other (more arithmetic) conditions, for which the equation $2^t = 2t$ holds, for $t \in \mathbb{N}$?

It appears to be true when $t = 2^s$, for some $s \in \mathbb{N}$.
The first equality is true: $\sqrt{2^{2t}}=\sqrt{2^{t+t}}=\sqrt{2^t\cdot2^t}=2^t$.

The second holds only for $t=1$ and $t=2$. Indeed, define for $t\in\Bbb R$ $$f(t)=\cfrac{2^{t}}{2t}$$ and $$g(t)=\ln f(t)=t\ln 2-\ln 2-\ln t$$
Now,
$$g'(t)=\ln2-\frac1t$$

We see that $g'$ vanishes only at one point, namely $t=1/\ln 2$. By Rolle's theorem, $g$ has no more than two zeros, and hence the equation $f(t)=1$ can not have more than two solutions.
my proof using differentiation


Let
$$f(t)=2^t-2t$$


  $$f(2)=2^2-2×2=4-4=0 \cdots  (1)$$


$$f'(x)=\ln(2) 2^t  -2$$


  $$f'(2)=4\ln(2) -2=0.772589\cdots(2) $$


$$f''(t)=(\ln(2))^2 2^t&gt;0$$
$$\text{for }x\in R$$


  $$\text{.: f'(t) is an increasing function for }x\in R$$


$$\text{but }f'(2)&gt;0$$


  $$\text{.:} f'(t)&gt;0 \text{ for }x\gt2$$


$$\text{.: }f(t)\text{ is an increasing function for }x\gt 2$$


  $$\text{but }f(2)=0$$


$$\text{.: }f(2)\gt 0 \text{ for }x\gt2$$


  $$\text{.:}2^t-2t \gt 0 \text{ for }x\gt2$$


$$\text{.:}2^t\gt 2t\text{ for }x\gt 2$$