Solution 1. The only such prime is $p=3$, when $2^3 + 1 = 3^2$. We consider the remainder of $2^k + 1$ on division by $9$, for arbitrary integers $k$ and first note that $2^6 = 64 \equiv 1 \pmod{9}$ and so
$$2^{k+6} + 1 \equiv 2^6 2^k + 1 \equiv 2^k + 1 \pmod{9}.$$
The statement of the question clearly does not hold when $p=2$ so we must consider only odd primes. In that case, we can determine the following table of $2^k + 1 \pmod{9}$ given $k \pmod{6}$:

If $3$ divides a perfect square, then so must $9$ and thus perfect squares can arise only when $k \equiv 3 \pmod{6}$ and in particular $k$ is a multiple of $3$. The only prime that is also a multiple of $3$, is $3$ itself.

Solution 2. We first note that $2^p + 1 = n^2$ implies that $n$ needs to be odd. The given equation can be written as $2^p = (n-1)(n+1)$ with both factors on the right hand side even, hence $\text{gcd}(n+1, n-1) = \text{gcd}(n+1, 2) = 2$. As $n-1$ and $n+1$ must both be powers of $2$, it follows that the smaller of them, $n-1$, must be equal to $2$. This shows that $n=3$ which leads to the unique solution $p=3$. The assumption that $p$ is prime was not needed.