Does the equation $k^k+1=a^2$ have any positive integer solutions?

Question

As the title of the question says, I want to know if the equation $k^k+1=a^2$ has any integer positive solutions, (which means k and a are both positive integers.)
I'd really appreciate it if someone could provide the following:

1. If the given solutions exists, please provide the minimal solution. Are the solutions infinite or finite?
2. If the given solutions doesn't exist, please provide your proof.

Answer 1

There are no positive integer solutions by Mihailescu's theorem.

With a less powerful hammer, Ke Zhao proved the nonexistence of solutions to $x^y+1=a^2$ in 1964. See the last reference in this answer.

Answer 2

$k^k+1$ itself has factors, one of which is $k+1$. These would have to be coprime, or a common divisor of 2, because the common factor between $(a^k+1)$ and $(k+1)$ would divide 2k.

Answer 3

$k^k+1=a^2\Rightarrow k^k=(a-1)(a+1)$

$\gcd(a-1,a+1)=1\text{ or }2$

Case 1: $\gcd(a-1,a+1)=2$. Then $k$ is even, and $k^k$ is a perfect square. The only two squares that differ by $1$ are $0,1$, but $0$ is not a positive integer.

Case 2: $\gcd(a-1,a+1)=1$. Then both $(a-1)$ and $(a+1)$ are perfect $k$th powers that differ by $2$. This can only occur when $k=1$, but $1^1+1=2$ which is not a perfect square.