Few days ago I found a question which was unfortunately put "on hold" and later it was even deleted. The question asked to solve the following exponential diophantine equation $$p^{p+1}+(p+1)^p=x^2$$ where $x$ is a positive integer and $p$ is a prime number. Since I like diophantine equations I found this one very interesting and I tried to solve it.
Here is what I did: we can assume that $p$ is odd because for $p=2$ there is no solution, so $p+1=2n$ and thus we can write $$(2n)^p=(p+1)^p=x^2-p^{2n}=(x+p^n)(x-p^n).$$
Now, from the original equation we get that $x$ is odd and $x\equiv \pm 1\pmod p$, then $\gcd(x+p^n, x-p^n)=2$, and therefore we can write $x+p^n=2^{\alpha}\cdot y$ and $x-p^n=2^{\beta}\cdot z$, with exactly one of $\alpha$ and $\beta$ being equal to $1$ because $\alpha+\beta\ge p\ge 3$. More precisely, if $n=2^r\cdot m$, then $$2^{p+rp}\cdot m^p=2^{\alpha+\beta}\cdot yz.$$
Hence $p(r+1)=\alpha+\beta$ and $m^p=yz$ with $\gcd(y,z)=1$, which means that $y=y_1^p$ and $z=z_1^p$, for some $y_1, z_1\in\Bbb{Z}^+$ such that $\gcd(y_1, z_1)=1$.
Sadly, I couldn't do anything else and I got stuck here until I remembered that the general equation $y^x+x^y=z^2$ was solved in this paper when $\gcd(x,y)=1$, $xy$ is even and $\min\{x,y\}>1$.
In our particular equation the required conditions are satisfied because $\gcd(p+1,p)=1$, $(p+1)p$ is even and $\min\{p+1,p\}\ge 3$, so according to that paper our equation doesn't have any solutions. The only problem that I found is that the equation $y^x+x^y=z^2$ was solved using bounds for linear forms in logarithms, which are beyond from being elementary.
So this is my question: is there a more elementary solution for the equation $p^{p+1}+(p+1)^p=x^2$ ? Thanks in advance for your answers.