Define the fully factored form (FFF) of an integer $n\ge 2$ to be the result of the following procedure:
Write the prime factorization of $n$ in standard form $\prod_{i=1}^{k} p_i^{n_i},$ where $p_1 < p_2 < ... < p_k$ are primes and the $n_i$ are positive integers, then do the same for each exponent $n_i,$ and continue doing this until only prime numbers or $1$s appear in the expression. Finally, remove all $1$s.
E.g., $\text{FFF}(2^{18}\cdot 3^{64})=2^{2\cdot 3^2}\cdot 3^{2^{2\cdot 3}}$.
Now consider the following infinite sequence of FFFs, in which each successive FFF is obtained from the previous FFF by adding $1$ to every prime number occurring in it:
$$\begin{align}&2\\ &3\\ &4=2^2\\ &3^3\\ &4^4=2^{2^3}\\ &3^{3^4}=3^{3^{2^2}}\\ &4^{4^{3^3}}=2^{2^{5\cdot 11}}\\ &3^{3^{6\cdot 12}}=3^{3^{2^3\cdot 3^2}}\\ &4^{4^{3^4\cdot 4^3}}= 2^{2^{10369}}\\ &3^{3^{10370}}=3^{3^{2\cdot 5\cdot 17\cdot 61}}\\ &4^{4^{3\cdot 6\cdot 18\cdot 62}}=2^{2^{40177}}\\ &3^{3^{40178}}=3^{3^{2\cdot 20089}}\\ &4^{4^{3\cdot 20090}}=2^{2^{149\cdot 809}}\\ &3^{3^{150\cdot 810}}=3^{3^{2^2\cdot 3^5\cdot 5^3}}\\ &4^{4^{3^3\cdot 4^6\cdot 6^4}}=2^{2^{5\cdot 227\cdot 252559}}\\ &3^{3^{6\cdot 228\cdot 252560}}=3^{3^{2^7\cdot 3^2\cdot 5\cdot 7\cdot 11\cdot 19\cdot 41}}\\ &\cdots \end{align}$$
Clearly the sequence must alternate between powers of $2$ and powers of $3$, and we notice that each of the first four terms is a tetration (i.e., it's an exponential tower of $2$s only or $3$s only).
Q: How to settle the question of whether a tetration appears in this sequence after $3^3$? Are there no more at all? Or do they occur infinitely often?
NB: If any more tetrations do occur, they must be towers of $3$s only, because no step of form $3^{3^x}\to 4^{4^y}=2^{2^{2y+1}}$ can result in a tower of $2$s only, as $2y+1$ is odd.
However, in the steps of form $2^{2^x}\to 3^{3^y}$, is there some number-theoretic argument against $y$ occasionally (though rarely) being a tower of $3$s only?
NB: Following are the exponents $x$ in the first $12$ terms of form $3^{3^x}$, suggesting that these exponents are always even, thus preventing any more tetrations (but how to prove this?):
2^2
2^3 * 3^2
2 * 5 * 17 * 61
2 * 20089
2^2 * 3^5 * 5^3
2^7 * 3^2 * 5 * 7 * 11 * 19 * 41
2^8 * 3^3 * 5 * 13 * 41 * 97 * 251
2^19 * 3^7 * 7^2 * 31 * 619 * 683 * 1489 * 1709 * 15420877 * 758647158424919209
2^5 * 3^2 * 5 * 7 * 11^2 * 59 * 6247 * 128703053 * 1678759855446808823464308071758681001
2^2 * 5^3 * 23 * 79 * 25541 * 197316931971016973669 * 137517302030623161523706659273
2^4 * 7 * 1117 * 1657 * 605660678957687521638971 * 741743082447608620588796259899
2^4 * 3^4 * 5 * 7 * 31 * 136859 * 811553 * 1101593 * 14558849 * 132381127 * 40222263624231587 * 7742819802333668771
Summarizing: If we let $f(x)$ be the result of incrementing all the primes in the full factorization of $x$, then the original sequence has $3^{3^x}\to 2^{2^{2f(x)+1}}\to 3^{3^{f(2f(x)+1)}}\to\cdots $. Since $2f(x)+1$ is always odd, $2^{2^{2f(x)+1}}$ can never be a tower of $2$s only, and if $f(2f(x)+1)$ is always even, then $3^{3^{f(2f(x)+1)}}$ can never be a tower of $3$s only, thus precluding any more tetrations at all.
It's already been shown that after the term $3^3$ the sequence alternates between terms of form $2^{2^x}$ with odd $x\ge3$, and terms of form $3^{3^y}$. Since $x\ge3$ is odd, no term of form $2^{2^x}$ can be a tower of $2$s only. Furthermore, in any step of type $2^{2^x}\to 3^{3^y}$, $y$ is the result of adding $1$ to each prime in the FFF of $x$. Since $x\ge3$ is odd, $x$ is a product of odd primes (i.e., $2$ may occur only in exponents of the FFF of $x$), so $y$ must be a positive even number; hence, $3^{3^y}$ cannot be a tower of $3$s only. Therefore, after the term $3^3$, no term can be a tetration.