$$p_n-p_{\lfloor n^{\frac{1}{m}}\rfloor^m}=0 \Rightarrow m=2$$
$$\quad\quad(\operatorname{tooth1})$$
$$p_n-p_{\lfloor n^{\frac{1}{m}}\rfloor^m}\not=0 \Rightarrow m\not=2$$
$$\quad\quad(\operatorname{tooth2})$$
$p_n-p_{\lfloor n^{\frac{1}{m}}\rfloor^m}=0$ if and only if $m=2$.
$$\quad\quad(\operatorname{tooth3})$$
If the first two teeth are true, then it is obvious that our third tooth also is true.
So my problem is reduced to seeking $(n,m)$ such that $m\not=0$ but $p_n-p_{\lfloor n^{\frac{1}{m}}\rfloor^m}=0$ as the necessary conditions for a counter example to the initial conjecture made with the stated teeth, therefore we can conclude it is false and no futher investigation is required, or a proof that such a counter example existing is an impossibility over the infinitude of primes for which the prime gap stated is taken, assuring us that the congruence of $p_n$ and $p_{\lfloor n^{\frac{1}{m}}\rfloor^m}$ modulo unity can only be true in the case that $m=2$, so here a I am looking for a means of determining these truth values and taking these considerations into account for another problem.