Is $28$ the only perfect number that is form of $n^n+1$?
I noticed that in the sequence of $n^n+1$, $28$ is the only perfect number that is in the list.
Using Pari GP, I checked the values of $n\leq10^{4}$, but it seems that $28$ is the only perfect number showed in the list.
I think that $n+1$ must be divisible by $4$ in order for $n^n+1$ be a perfect number.
Note that $n\in \mathbb{Z}^{+}$.
I can give you an answer for even perfect numbers.
Lemma: If $a,b\in\mathbb N$ such that $a^b\equiv -1\pmod{2^k}$ for $k\geq 1,$ then $a\equiv -1\pmod{2^k}.$
Proof: By induction on $k.$
I leave it to you to show $k=1.$
If true for $k,$ and $a^b\equiv -1\pmod{2^{k+1}}$ then $a^b\equiv -1\pmod{2^k}$ and hence, by the induction hypothesis, $a\equiv -1\pmod{2^k}.$
This means that $a\equiv -1$ or $a\equiv 2^k-1$ $\pmod{2^{k+1}}.$
If $a\equiv 2^k-1\pmod{2^{k+1}},$ then: $$a^2\equiv 2^{2k}-2\cdot 2^k+1\equiv 1\pmod{2^{k+1}}.$$ So it is not possible for $a^b\equiv -1\pmod{2^{k+1}},$ it only takes values $1$ and $a.$
The even perfect numbers are of the form $m_k=2^{k}(2^{k+1}-1)$ where $2^{k+1}-1$ is prime.
We will show that, for $k\geq 3,$ $m_k$ can't be written as $n^n+1,$ whether $m_k$ is perfect or not.
If $n^n+1=m_k,$ then, in particular, $n^n\equiv -1\pmod{2^k},$ and thus $n\equiv -1\pmod{2^k}$ by the lemma.
But then $n\geq 2^{k}-1>2^{k-1}$ so $n^n>2^{(2^k-1)(k-1)}.$
And $m_k<2^{2k-1}.$
So all that is left is to show $(2^k-1)(k-1)> 2k-1$ for $k\geq 3.$ (When $k=2,$ the two values are equal, yielding your one example.)
Now, $2^k\geq k+1,$ so it is enough to show $k(k-1)> 2k-1$ or $k(k-3)\geq -1$ for $k\geq 3.$ But that is trivially true.
So we've proven it for all even perfect numbers.
The question is whether we know enough about possible odd perfect numbers to eliminate those cases, too. There are a lot of theorems of the form: "If $m$ is an odd perfect number, then ..." But I don't know them well enough to say if we can deduce this.
Certainly, it would mean $n$ must be even and hence $m\equiv 1\pmod{2^n}.$
It also means $m=1+c^2$ where $c=n^{n/2}$ This means the prime factors of $m$ must all be $\equiv 1\pmod 4.$