It is given as an exercise question in the book titled "An Introduction to Essential Algebraic Structures" By Martyn R. Dixon, Leonid A. Kurdachenko, Igor Ya Subbotin.
I feel there should be infinite number of values.
Manually, only one values found by me:
(i) $x=2 \implies 5$.
In $\Bbb Z$, an element $p$ is prime iff it is irreducible, that is, iff the only divisors of $p$ are $p$ and $\pm 1$.
If you factorise your quadratic, you get $x^2 + 2x - 3 = (x+3)(x-1)$. Applying the above criterion, you need one of these factors to be $1$ (since in the question you are looking in $\Bbb Z^+$). Indeed, this means that $x = 2$ gives $(2 + 3)(2 - 1) = 5$, and since this is the only positive integer which makes one of these factors $1$, this is the only $x \in \Bbb Z^+$ making $x^2 + 2x - 3$ prime.