This is a continuation of the problem asked here. I made some edits to the original problem, but enough changes were made to warrant a new question.
The goal is, given a value of $p$, find the maximum value of $x$ that satisfies $$ x^{\frac{p^2+p+1}{p^2}}-x^{\frac{p+1}{p}} < \log(p) $$ given $p\in\mathbb{Z}\geq 3$.
For purposes of usage, a lower bound of $x<x_{max}$ would be signifigantly more useful, although an upper bound may potentially be interesting.
In their answer, user Claude Leibovici showed that for sufficiently large $p$, $x_{max}\approx \frac{1}{2}p^2$.
Below is my work that I appended to the bottom of the original question. It provides a lower bound estimate that turns out to be signfigantly lower than the true value (but not too bad) $$ e^{W(p^{k+1}-1} < x_{max} $$
We can also think of this as the inequality of two continuous functions $$ f(x;k,p) = x^{\frac{p^{k+1}-1}{p^k(p-1)}} $$ $$ g(;a,k,p) = a + k\log(p). $$
Note that $a$ is a constant derived from the problem that guarantees that $f(x; 1,p) < g(;a,1,p)$ for a given value of $x$. The initial question formulation was asking to find the largest value of $x$ that satisfied the following inequality $$ f(x;2,p) - f(x;1,p) < g(;a,2,p) - g(;a,1,p) $$ with the goal of if we know a priori that $f(x;1,p) < g(;a,1,p) $ we can then imply that $f(x;2,p) < g(;a,2,p).$
We can differentiate both sides of the second inequality with respect to $k$ to get $$ x^{\frac{p^{k+1}-1}{p^k(p-1)}} \log(x)\left(\frac{\log(p)}{p^k(p-1)}\right) < \log(p) $$ for which we want to find a value of $x$ max that makes this inequality true. Rearranging we get that $$ x^{\frac{p^{k+1}-1}{p^k(p-1)}} \log(x) < p^k(p-1) $$ setting $x = e^z$ and just using some temp placeholder variables we can say that $$ e^{az}z < b $$ which has solution $$ z < W(ab) \implies \log(x) < W(p^{k+1}-1) $$ so our max value is $x=e^{W(p^{k+1}-1)}$
Since the derivative of the left hand side seems to be decreasing on $[1,2]$, is it sufficient to say that if $x$ is less than our bound here it should also be less at the next point?
This may just be me rambling but maybe it is helpful to the task
For this answer, I shall refer to what I wrote for this previous question.
Considering the problem of the zero of function $$f(x)=x^a-x^b -\log(p) \quad \text{where} \quad a=\frac{p^2+p+1}{p^2}\quad \text{and} \quad b=\frac{p+1} p$$ $$f'(x)=a x^{a-1}-b x^{b-1} \quad \text{and} \quad f''(x)=a(a-1) x^{a-2}-b(b-1) x^{b-2}$$ I suggested as an initial estimate $$x_0=\frac{p^2\log (p)}{W\left(p^2 \log (p)\right)}$$ and Newton iterates will be $$x_{n+1}=x_n-\frac{x_n^a-x_n^b -\log(p) }{a x_n^{a-1}-b x_n^{b-1} }$$ This converges very fast.
Using as before for a test, $p=123.456$ and working with unlimited precision we have $$\left( \begin{array}{cc} n & x_n \\ 0 & 8150.36957189589 \\ 1 & 7636.04252044307 \\ 2 & 7634.23710338033 \\ 3 & 7634.23708015417 \\ 4 & 7634.23708015417 \end{array} \right)$$ which shows that we know the analytical solution of the equation.
What has been checked is that, as a function of $p$, $f(x_n)$, $f'(x_n)$ and $f''(x_n)$ are always positive. So, by Darboux theorem, $x_n$ is an overestimate of the solution. So, to satisfy the inequality, $\color{red}{\text{a lower bound is } \lfloor x_n\rfloor ~~\forall n \ge 2}$
Now, let us consider the implicit function $$g(x,p)=x^a-x^b -\log(p)=0$$ $$\frac{\partial g(x,p)}{\partial p}=\frac{x^b \log (x)}{p^2}-\frac{(p+2) x^a \log (x)}{p^3}-\frac{1}{p}$$ $$\frac{\partial g(x,p)}{\partial x}=a x^{a-1}-b x^{b-1}$$ and then, at solution, $$\frac{dx}{dp}=-\frac{\frac{\partial g(x,p)}{\partial p} } {\frac{\partial g(x,p)}{\partial x} }$$
Numerical analysis shows that $$\frac{dx_n}{dp}\sim p-2$$
For $n=3,4$ a quick and dirty linear regression $\frac{dx_n}{dp}=a+b\,p$ gives with $R^2=0.999977$ $$\begin{array}{clclclclc} \text{} & \text{Estimate} & \text{Standard Error} & \text{Confidence Interval} \\ a & -1.86836 & 0.05913 & \{-1.98572,-1.75099\} \\ b & +1.03026 & 0.00101 & \{+1.02825,+1.03227\} \\ \end{array}$$
Edit
A much better estimate of the solution is given by the first iterate of Halley method $$x_1=x_0+\frac{2 f(x_0) f'(x_0)}{f(x_0) f''(x_0)-2 f'(x_0)^2}$$ with the same $x_0$ as before.
For $p=123.456$, $x_1=7634.28$.
More spectacular would be the first iterate of Householder method $$x_1=x_0+\frac{3 f(x_0) \left(f(x_0) f''(x_0)-2 f'(x_0)^2\right)}{f(x_0)^2 f'''(x_0)+6 f'(x_0)^3-6 f(x_0) f'(x_0) f''(x_0)}$$
Update
Just out of curiosity, I searched the maximum value of the ratio $$R_n=\frac{x^a} {x^b +\log(p)}$$ for Newton successive iterates $$\left( \begin{array}{cc} n & R_n \\ 0 & 1.11721588695 \\ 1 & 1.01728777375 \\ 2 & 1.00039248588 \\ 3 & 1.00000020331 \\ 4 & 1.00000000076 \end{array} \right)$$