Peculiar locations of the root and the maximum of $(x+1)^{x+1}-x^{x+2}$ Related to some other problems, I got interested in this function:
$$(x+1)^{x+1}-x^{x+2}$$
Its root is very close to $\pi$. This is Mathematica code that finds the root:
NSolve[Power[x + 1, x + 1] - Power[x, x + 2] == 0, x, Reals]


{{x -> 3.14104}}

($\pi \simeq 3.14159$)
Moreover, the location of its maximum is very close to $e$:
FindMaximum[Power[x + 1, x + 1] - Power[x, x + 2], {x, 2.65, 2.78}]


{20.0645, {x -> 2.70965}}

($e\simeq 2.71828$)
I find these facts almost disturbing.

Is there a deeper mathematical explanation for such "close but not quite" behavior?

I tried various ways of approximation of this function (similar to this), but don't have enough math knowledge and background to get any result that makes sense.
Here is the graph of the function:

 A: 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.
