Inspired by Spy Koni

$\begin{aligned} &n^{\left(n^n\right)}={\left(n^n\right)}^n \\ &n^{n^n}=n^{n^2} \\ &n^{n^n}-n^{n^2}=0 \\ &n^{n^2}\left(n^{n^n-n^2}-1\right)=0 \\ \end{aligned}$

Case 1: $n^{n^2} = 0$

Clearly, $n \neq 0$ , because $0^0$ is not defined.
Since, $n$ is a non-zero integer, $n^2$ is a positive integer.

Now, $n = 0$ which has no solutions.

Case 2: $n^{n^n-n^2}$ = 1

If $n^n - n^2 = 0$ , then $n^2\left(n^{n-2}-1\right)=0$ .
Since, $n \neq 0$ , $n^{n-2} = 1$ .
Now if $n = 2$ , we have a solution.
Now if $n > 0, n \neq 2$ , we have $n = 1$ , which is also a solution.
Now if $n < 0$ , $n$ must be even because both sides of the equation must have the same sign.
Now, there exists integer $k$ such that $n = 2k$ .
Now $n^{2k - 2}=1 \Rightarrow n^2 = 1$ , which has solution $n = -1$ .
However, $n$ must be even so it is not a solution.

Thus, if $n^n - n^2 = 0$ , $n = 1$ or $n = 2$ , which are also solutions to the original equation.

Now if $n > 0, n \neq 1, n \neq 2$ , we must have $n = 1$ , which has no solutions.

Now if $n < 0$ , we must have $2 \mid n^n - n^2$ , which means that $n^n - n^2$ must be an integer, which means that $n^n$ must be and integer, because $n^2$ is an integer.
However, if $n < -1$ , $n^n = \dfrac1{n^{-n}}$ is not an integer, which means that $n = -1$ is the only candidate and it indeed is a solution.

Hence, in case $2$ , the solutions are $n = -1, n = 1, n = 2$

Conclusion:

Now, combining the cases we have $n = -1, n = 1$ and $n = 2$ as the only solutions.

Hence, the answer is $\boxed{3}$ .

1 and -1 can be easily seen to satisfy the equation (Note 0 cannot be a solution as 0^0 is not defined)

Assuming x to be + ve , For the other solution take log on both sides. This gives us

Log (x^x^x)= log (x^x^2)

x^x log(x)= x^2 log (x)
{ log (a^b)= b log(a) }

x^x=x^2

Thus x=2 or 1

(And the trivial solution of -1 to the original equation )

So there are 3 integer solutions to x