Peculiar locations of the root and the maximum of $(x+1)^{x+1}-x^{x+2}$

This is not an answer, but too long for a comment.

The most interesting part is the root location, so I'm going to clarify my comment about the series expansion.

The equation for the root can be transformed to have the form:

$$x=\left(1+\frac{1}{x} \right)^{1+x}$$

Consider the function:

$$f(x)=\left(1+\frac{1}{x} \right)^{1+x}$$

Let's change the variable $1/x \to y$, and expand the resulting function around $y \to 0$ up to third order terms:

$$\left(1+y \right)^{1+\frac{1}{y}}=(1+y)\exp \left( \frac{1}{y} \ln (1+y)\right)=(1+y)\exp \left( 1-\frac{y}{2}+\frac{y^2}{3}-\frac{y^3}{4}+\dots\right) =$$

$$= e(1+y) \left(1+\left(-\frac{y}{2}+\frac{y^2}{3}-\frac{y^3}{4} \right)+\frac{1}{2} \left(-\frac{y}{2}+\frac{y^2}{3} \right)^2+\frac{1}{6} \left(-\frac{y}{2} \right)^3 \right)+O(y^4)=$$

$$=e \left(1+\frac{y}{2}-\frac{y^2}{24}+\frac{y^3}{48}\right)+O(y^4)$$

So finally we have an approximation:

$$x \approx e \left(1+\frac{1}{2x}-\frac{1}{24x^2}+\frac{1}{48x^3}\right)$$

Substituting $\pi$ on the right side we get:

$$x \approx e \left(1+\frac{1}{2\pi}-\frac{1}{24\pi^2}+\frac{1}{48\pi^3}\right)=3.14126\dots$$

I don't see any deeper reason for $\pi$ to be a good approximation for the root of this function. It seems to be a coincidence.

There should be nothing surprising about the maximum, because this function is closely related to $e$.

But it turns out to be tricky. Let's find the fist derivative of the function:

$$g(x)=(1+x)^{1+x}-x^{2+x}=e^{(1+x)\ln(1+x)}-e^{(2+x)\ln(x)}$$

$$g(x)'=(1+\ln(1+x))e^{(1+x)\ln(1+x)}-\left(1+\frac{2}{x}+\ln(x) \right) e^{(2+x)\ln(x)}$$

The equation for the extremum is:

$$g(x)'=0$$

Which can be transformed to:

$$\left(1+\frac{1}{x} \right)^{1+x}=\frac{2+x(1+\ln x)}{1+\ln(1+x)}$$

If we substitute $e$ on the right hand side, we get:

$$\frac{2+x(1+\ln x)}{1+\ln(1+x)}=\frac{2(1+e)}{1+\ln(1+e)}=3.215\dots$$

On the left hand side we get (note that for any $x$ we know that $\left(1+\frac{1}{x} \right)^{1+x}>e$):

$$\left(1+\frac{1}{x} \right)^{1+x}=\left(1+\frac{1}{e} \right)^{1+e}=3.205\dots$$

Again, I'm not sure if we can make any conclusions from the fact that the solution is close to $e$.

But I hope these calculations help the OP get a better understanding about this function.