In a field such as $\:\mathbb Q\:,\ \mathbb R\:,\ \mathbb C\:,\:$ we have $\rm\ x^2 = 1\ \iff\ (x-1)\ (x+1) = 0\ \iff\ x = \pm 1\:.\: $ In rings that are not fields there can be more than two square-roots, e.g. modulo $15$ there are two additional roots $\rm\ (\pm\:4)^2\equiv 1\ (mod\ 15)\:.\:$ In some contexts authors define single-valued square-root functions that uniformly select one of the roots, e.g. the non-negative root, or principal branch, etc. In ring theory one can adjoin a generic square root of one $\rm\:s = \pm 1\:$ to a ring $\rm\:R\:$ where $\:2\:$ is invertible as follows $\rm\ R[s]\ \cong R[x]/(x^2-1)\ \cong R/(x-1) \times R/(x+1)\ \cong R^2\:,\ $ via $\rm\ f(x)\to (f(1),\ f(-1))\:;\:$ informally, do arithmetic in two parallel universes (rings), one with $\rm\ s = 1\ $ and one with $\rm\ s = -1\:.$
As to your second question concerning why there aren't other numbers that are their own square root, note that $\rm\ \sqrt{x} = x\ \Rightarrow x = x^2\ \Rightarrow\ x\ (x-1) = 0\ $ so $\rm\:x = 0\:$ or $\rm\:x = 1\:$ in a field (or domain). Elements satisfying $\rm\ x^2 = x\ $ are known as idempotents. Idempotents that are nontrivial ($\ne\: 0,\:1$) are intimately connected with nontrivial factorizations. In the above example $\:\mathbb Z/15\:,\:$ the integers modulo $\:15\:,\:$ we have the factorization $\ \mathbb Z/15\ \cong \mathbb Z/3 \times \mathbb Z/5\:,\:$ arising from the idempotent $\rm\:6\:,\:$ i.e. $ (0,1)\in \mathbb Z/3 \times \mathbb Z/5\:.\:$ Note that if $\rm\:e\:$ is idempotent then $\rm\:(1-2\:e)^2 =\: 4\:(e^2-e)+1 = 1\:,\: $ hence $\rm\:2\:e-1\:$ is a square-root of $1\:;\:$ it is a nontrivial square root $\rm\:2\:e-1\:\ne\: \pm 1\:$ if $2$ is cancellable. In our example $\rm\:e=6\:,\:$ so $\rm\:1-2\:e \equiv 4\:$ which does indeed square to $\rm\:1\ (mod\ 15)\:.\:$ We can quickly factor $15$ from this nontrivial square root $\rm\:s\:,\:$ namely $\rm\:gcd(15,s-1) = gcd(15,3) = 3\:;$ $\rm\ gcd(15,s+1) = gcd(15,5) = 5\:.\:$ In fact many integer factorization algorithms work by searching for nontrivial square-roots. This way of factoring by simple gcd computations works more generally. One can quickly factor $\rm\:n\:$ given any polynomial with more roots $\rm\:(mod\ n)\:$ than the degree of the polynomial - see my post here.