In general, if $Q(x,y,z)$ is a nonsingular ternary quadratic form with integral coefficients, then the integral solutions of $Q(x,y,z)=n$ fall into finitely many orbits of the integral automorphism group of $Q$, and these orbits can be effectively determined. For your case there is a shortcut as follows.

Write $a=x+y$, $c=x-y$, $b=z$, then your equation becomes $b^2-ac=\mp 1$. In the language of binary quadratic forms, this means that the form $au^2+2buv+cv^2$ has discriminant $\mp 4$. From classical theory it follows that after an invertible linear change of variables $u'=pu+qv$, $v'=ru+sv$ the form becomes $u'^2\pm v'^2$, i.e. there are $p,q,r,s\in\mathbb{Z}$ such that $ps-qr=1$ and
$$ au^2+2buv+cv^2=(pu+qv)^2\pm(ru+sv)^2. $$
Equating coefficients,
$$ a=p^2\pm r^2,\quad c=q^2\pm s^2,\quad b=pq\pm rs, $$
i.e.
$$ x=(p^2\pm r^2+q^2\pm s^2)/2,\quad y=(p^2\pm r^2-q^2\mp s^2)/2,\quad z=pq\pm rs. $$
To summarize, all integer solutions of $x^2-y^2-z^2=\pm 1$ are of this form, with integers $p,q,r,s$ satisfying $ps-qr=1$. Using congruences one can easily distinguish integer solutions from half-integer ones.

As references, I suggest Rose: A course in number theory, and Cassels: Rational quadratic forms.

Pell equation, $$q^2-17p^2=-1$$ So we can actually make both the $Y,Z$ of your first eqn as squares. – Tito Piezas III Jan 3 '15 at 6:37