We show that $7$ is the sole integer satisfying the required conditions. If $a \in \mathbb{Z}_n$ and $a^2 - a + 2 = 0$ in $\mathbb{Z}_n$, then $(1-a)^2 - (1-a) + 2 = a^2 - a + 2 = 0$ in $\mathbb{Z}_n$, so uniqueness forces $a = 1-a$, i.e., $2a = 1$. In particular, $2$ is invertible in $\mathbb{Z}_n$ and $a = 2^{-1}$. Hence $2^{-2} - 2^{-1} + 2 = 0$, or $1 - 2 + 8 = 0$, i.e., $7 = 0$ in $\mathbb{Z}_n$. Consequently, $n$ divides $7$, and since the latter is prime, it follows that $n = 7$. It is readily checked that $a = 2^{-1} = 4$ is the unique element of $\mathbb{Z}_7$ satisfying $a^2 - a + 2 = 0$.