Question on partial answer of the generalized Catalan's conjecture (case $n =2$) Edited question.

Let $n,m\in\Bbb N$ be integers greater or equal to $2$. If $3\leq m<n$ then there is no $m,n$ such that $m^n -n^m =2$. If $n<m$ and $m$ is prime then there is no $m,n$ such that $m^n-n^m =2$. If $m = 2$ then there is no $n$ such that $m^n- n^m = 2$.

Original question.

Let $n\geq 2$ be an integer and $p$ be a prime number. Then there is no $p,n$ such that $p^n - n^p =2$.

False Proof : Consider the case when $p$ is odd prime. Then
by FLT, $n^{p-1}\equiv 1\bmod p$ so $n^p = p^n-2 \equiv n\bmod p$. $\color{#C00}{\text{Then $n^{p+1} = (p+1)^n -2\equiv n\bmod p+1$ so $n^p\equiv 1\bmod p+1$}}$. $p^n - 2 = (p+1)m+1$ for some $m$ so that $p^n =(p+1)m+3 = pm+m+3$. Hence, $p(p^{n-1}-m) = m+3 = pk_0$ for some integer $k_0$ so that $m = pk_0-3$. Then $p(p^{n-1}-pk_0+3) = pk_0$ so $p^{n-1}-pk_0= k_0-3$. Since, $p(p^{n-2}-k_0) = k_0-3$ and letting $k_0 = pk_1+3$, we get $p(p^{n-2}-pk_1-3) = pk_1$ i.e. $p^{n-2}-pk_1 = k_1+3$. Hence inductively, we conclude that $p(1-k_N) = k_N\pm 3$ for some large $N$. This gives the contradiction.
Rough idea given by @robjohn ♦: We want to show that $p^n-n^p=2$ is impossible, so if $n=kp-2$, consider $p^{kp-2}-(kp-2)^p$
Dividing both sides by $(kp)^p$, we have $\frac{p^{(k-1)p}}{k^pp^2}-\left(1-\frac2{kp}\right)^p$.
If $k=1$, this is $\frac1{p^2}-\left(1-\frac2{p}\right)^p$ which is less than $0$. If $k\ge2$, then $\left(\frac{p^{k-1}}{kp^{2/p}}\right)^p-\left(1-\frac2{kp}\right)^p$ is greater than $0$ and multiplied by $(kp)^p$ is much bigger than $2$.

This is a particular case of $m^n = n^m+2$ with $m,n\geq 2$. First note if both numbers are even then $m^n$ and $n^m$ are multiples of $4$, so we have $m,n \geq 3$. Let $n = m+k$ with $k\geq 1$. ( You can see that if $m$ is the larger one it wont work because the left side will be smaller than the right).
Taking logs gives $n\log(m) = m\log(n) + \epsilon$ where $\epsilon$ is the error term $\log(n^m+2) - \log(n^m) = \int \limits_{n^m}^{n^m+2} \frac{1}{x} dx \leq \frac{2}{n^m}$.
$n \log(m) - m\log(n) = (m+k)\log(m) - m(\log(m+k))$.
Differentiating with respect to $k$ we get $\log(m) - \frac{m}{m+k}$ so the function is increasing in $k$ in $(1,\infty)$.
Setting $k=1$ yields $(m+1)\log(m) - m\log(m+1) = \log(m) - m(\log(m+1) - \log(m))$ where $\log(m+1) - \log(m) = \int\limits_{m}^{m+1} \frac{1}{x}dx \leq  \frac{1}{m}$. It follows that $\epsilon$ is at least $\log(m) - 1 \geq \log(3) - 1$.
Hence we must have $\log(3) -1 \leq \frac{2}{n^m}$ or $n^m \leq \frac{2}{\log(3)-1}<21$.
$4^3$ is already $64$ so there is no candidate we must check.
