It seems reasonable to expect that 29 is the only prime, just because primes get rare as numbers get big, so we'd expect almost all of these numbers to be composite.

That is, heuristically, the prime number theorem gives us the "probability" that a number $n$ is prime is about $\frac{1}{\ln(n)}$.

Therefore, the expected number of primes of this form should be
$$
\sum_p \frac{1}{\ln(p^p + 2)} < \sum_p \frac{1}{p \ln(p)}
$$
where the sum is over all odd primes.
This latter sum converges, as shown in $\sum_{p \in \mathcal P} \frac1{p\ln p}$ converges or diverges?,
to roughly 0.92 (the sum over all primes is less than 1.64; the sum over all odd primes omits the term $\frac{1}{2\ln(2)}$.)

So, we'd "expect" to find about one prime of this form, and we have.

Of course, this isn't a proof; it assumes that numbers of the form $p^p + 2$ behave "randomly" with respect to prime factorization, rather than the factors having some special properties.

(Note that applying the same argument to numbers of the form $n^n + 2$, without the requirement that $n$ be prime, gives an expected number of primes approximately $\sum \frac{1}{n\ln(n)}$, which is infinite. Indeed, there are already 5 examples before 2000, at $n = 0, 1, 3, 737, 1349$.)

Compare Hardy and Littlewood's Conjecture E in Some problems of ‘Partitio numerorum’; III: On the expression of a number as a sum of primes:

There are infinitely many primes of the form $m^2 + 1$. The number
$P(n)$ of such primes less than $n$ is given asymptotically by
$$P(n) \sim C \frac{\sqrt{n}}{\ln(n)}$$
where
$$C = \prod_{p=3}^\infty \Biggl(1 - \frac{1}{p-1}\biggl(\frac{-1}{p}\biggr)\Biggr).$$

Also Conjecture K, which gives an aysmptotic number of primes of the form $m^3 + k$ for any fixed $k$ (except when $k$ is a cube).