Not a full answer, but some observations regarding the examples you have found and 2 new examples:
Consider the following examples:
$$
4^{16}\cdot 16^4 = 8^8\cdot 8^8
$$
$$
8^{144}\cdot 144^8 = 16^{96}\cdot 96^{16}
$$
$$
16^{144}\cdot 144^{16} = 32^{96}\cdot 96^{32}
$$
There is a clear recurrency here, in particular they are of the form:
$$
(2^{k})^{a}\cdot a^{2^k} = (2^{k+1})^b\cdot b^{2^{k+1}}
$$
and I have found two more! Namely:
$$
32^{1600}\cdot 1600^{32}=64^{1280}\cdot 1280^{64}
$$
$$
64^{1600}\cdot 1600^{64} = 128^{1280}\cdot 1280^{128}
$$
The pattern is now clear, and we can study it in order to generalize a possible new family of solutions. In particular I note that:
- If $k$ is odd then $a$ and $b$ satistifes both the "consecutive"
equations $$ (2^{k})^{a}\cdot a^{2^k} = (2^{k+1})^b\cdot b^{2^{k+1}}
$$ $$ (2^{k+1})^{a}\cdot a^{2^{k+1}} = (2^{k+2})^b\cdot b^{2^{k+2}}
$$
- Furthermore $\frac{a}{b} = 2$ in the first equation , $\frac{a}{b} =
\frac{3}{2}$ in the second and third, $\frac{a}{b} = \frac{5}{4}$ in
the fourth and fifth..
- And if $(a,b)$ is a solution to the "$k$-th and $k+1$-th equations,
then they also satisfy: $$ b^2 = a\cdot
(2^{k})^{(\frac{a-b}{2^{k}})}\cdot 2^{-(\frac{b}{2^{k}})} $$ $$ b^2
= a\cdot (2^{k+1})^{(\frac{a-b}{2^{k+1}})}\cdot 2^{-(\frac{b}{2^{k+1}})} $$
For example, in the last two $(a,b) = (1600,1280)$ and:
$$
1280^2 = 1600\cdot (2^{5})^{(\frac{1600-1280}{2^{5}})}\cdot 2^{-(\frac{1280}{2^{5}})} = 1638400
$$
$$
1280^2 = 1600\cdot (2^{6})^{(\frac{1600-1280}{2^{6}})}\cdot 2^{-(\frac{1280}{2^{6}})} = 1638400
$$
Now, of course it seems that for $k = 7$ and $k = 8$, $\frac{a}{b}$ should be $\frac{9}{8}$, but zero positive integer solutions to
$$
b^2 = a\cdot(2^{7})^{(\frac{a-b}{2^{7}})}\cdot 2^{-(\frac{b}{2^{7}})}
$$
using $\frac{a}{b} = \frac{9}{8}$. I have also tried other fractions but I just can not found other solutions. I have also brute forced the $k=7$-th equation and didn't found any solution for $a,b < 25000$.
EDIT 1:
Dividing the consecutive equations ($b^2 = \dots$) we obtain that for odd $k$:
$$
(1-k)a+kb = 0
$$
where $(a,b)$ is the solution to the $k$-th equation.
For example $(1-5)\cdot 1600 + 5\cdot 1280 = 0$.
EDIT 2
Using EDIT1 we know that the solution $(a,b)$ of the $7$-th equation has to satisfy:
$$
(1-7)a+7b = 0 \implies 7b-6a = 0
$$
and on the other hand:
$$
b^2 = a\cdot(2^{7})^{(\frac{a-b}{2^{7}})}\cdot 2^{-(\frac{b}{2^{7}})} = a\cdot 2^{\frac{7}{128}a-\frac{8}{128}b}
$$
which implies that $\frac{7}{128}a-\frac{8}{128}b = n\in\mathbb{N}$. The system of equations:
$$
\begin{cases}
7b-6a = 0 \\
7a-8b = 128n
\end{cases}
$$
has solutions given by $a = 896n$, $b = 768n$. Substituing back we obtain:
$$
(768n)^2 = 896n \cdot 2^{n}
$$
which has.. no integer solutions. Very disappointing, but maybe there are no more solution of this form :(
The following code in Mathematica
f[k_] := Reduce[{-(k - 1)* x + k* y == 0, k* x - 1 (k + 1)*y == 2^k *n}, {x, y, n}, Integers]
finds for odd values of $k$ the possible value of $a$ and $b$ as a function of $n$. Substituing back we can check with the script below if there exists $n\in\mathbb{N}$ that satisfy the equation.
NSolve[(b)^2 == a*2^n, {n}, PositiveInteger]
where instead of $(a,b)$ one should substitute the values obtained before.
EDIT 3
The solutions of the previous system are of the form:
$$
(a,b) = (nk 2^k, n(k-1) 2^k)
$$
for every odd $k$. Using the following code we can see that it seems that there are no integer solution to the equation $b^2 = a2^n$ for $k \geq 7$:
g[k_] := NSolve[((k - 1)*2^k*n)^2 == k*2^k*n*2^n, {n}];
Table[g[2*k + 1][[3]], {k, 1, 100}]
