direct proof $ x^2 \pm 1$ is not a perfect cube for integer $ x\geq 4$ by direct I mean wthout using any form of catalan's conjecture.
Since all even cubes are multiples of $8$ so they are multiples of $4$. Therefore if the square is odd and smaller than the cube  it has to be congruent to $3 \pmod 4$. And this is not possible since squares are congruent to $0$ or $1 \pmod 4$. So this gets rid of $1$ of those cases. However I feel that is not a good way to approach the problem.
 A: [The last case of $x^2-1=y^3$ is incomplete. Sorry.]
Let's look at the case
$$x^2+1=y^3.$$
As the l.h.s. is never divisible by four, the OP observed that we must have $y$ odd and $x$ even.
In the ring of Gaussian integers $\mathbb{Z}[i]$ the l.h.s. factors as $x^2+1=(x+i)(x-i)$. Here
$$
\gcd(x+i,x-i)=\gcd((x+i)-(x-i),x-i)=\gcd(2i,x-i)=1,
$$
because the norm of $x-i$ is odd, and it thus cannot be divisible by the sole prime factor $1\pm i$ of $2i$. Uniqueness of factorization in $\mathbb{Z}[i]$ then implies that
$$
x+i=u(a+bi)^3
$$
for some unit $u$ and a factor $a+bi$ of $y$. As $u$ is a cube itself, it can be absorbed into $a+bi$, so we need to study the equation
$$
x+i=(a+bi)^3=(a^3-3ab^2)+i(3a^2b-b^3).
$$
Comparing the imaginary parts gives us the equation (in $\mathbb{Z}$)
$$
1=b(3a^2-b^2).
$$
Factorization in the ring of integers then allows us to conclude that
$$
3a^2-b^2=b=\pm1.
$$
We easily see that the only possibility here is $b=-1$, $a=0$, which gives $x=0$, $y=1$ as the only solution.

Turning focus on 
$$
x^2-1=y^3.
$$
The l.h.s. factors as $(x-1)(x+1)$, so it is natural to split according to
whether $\gcd(x-1,x+1)=1$ or $2$, i.e. treating the cases of an even and odd $x$
separately.
If $x$ is even, then $\gcd(x-1,x+1)=1$, so uniqueness of factorization implies that $y=ab$ in such a way that $x-1=b^3$ and $x+1=a^3$. This means that
$$
2=(x+1)-(x-1)=a^3-b^3=(a-b)(a^2+ab+b^2),
$$
where both factor on the r.h.s. are positive, because the quadratic form is
positive definite. Here $a^2+ab+b^2\neq2$ for all integers $a,b$ (something that anyone who has played with the norm of Eisensteinian integers will know), so we are left with the system
$$
a-b=2\qquad\text{and}\qquad a^2+ab+b^2=1.
$$
The only solution of this is easily seen to be $a=1$, $b=-1$, which yields $x=1$, $y=0$.
That leaves the case $2\nmid x\Leftrightarrow 2=\gcd(x+1,x-1)$. In this case we have two possibilities
$$x+1=4a^3,\qquad x-1=2b^3$$
and
$$x-1=4a^3,\qquad x+1=2b^3.$$
These lead us to study the diophantine equations
$$
2a^3=b^3\pm1.
$$
This calls for the ring of Eisensteinian integers (another UFD). So let's denote a primitive third root of unity by $\omega=(-1+\sqrt{-3})/2$. We also record the relation $\omega^2+\omega+1=0$.
The r.h.s. factors over $O=\mathbb{Z}[\omega]$ as
$$
2a^3=b^3\pm1=(b\pm 1)(b\pm\omega)(b\pm\omega^2).
$$
Here we get a semblance of order from the observation that the numbers $1,\omega$
and $\omega^2$ are the representatives of the non-zero cosets of $2$ in the ring $O$. The l.h.s. is divisible by $2$ (inert in $O$), so one of the factors on the r.h.s. needs to be also. But the factors on the r.h.s. fall into different cosets of $2O$, so exactly one of them will be divisible by two. The ring $O$ has six units only two of which are cubes, so the number of cases blows up.
They all look like
$$
\begin{cases}b\pm1=2u_1a_1^3\\b\pm\omega=u_2a_2^3\\b\pm\omega^2=u_3a_3^3,\end{cases}
$$
where $u_i$ are units of $O$ and $a_i$ are arbitrary elements of $O$, $i=1,2,3,$
and we have the liberty of moving that extra factor $2$ from one equation to another.
A cube in the ring $O$ looks like ($c_1,c_2$ are rational integers)
$$
(c_1+c_2\omega)^3=(c_1^3-3c_1c_2^2)+3c_1c_2(c_1-c_2)\omega.
$$
Here the coefficient of $\omega$ is always divisible by three. This allows us to exclude elements such like $b\pm\omega$ and $b\pm\omega^2$ from being cubes in $O$. That forces $u_2$ and $u_3$ to be non-real. We can actually identify the unit up to sign, because the quotient ring $O/2O$ is a finite field of four elements, and thus all cubes of elements of elements in $O\setminus 2O$ are in the coset $1+2O$.
But I cannot allot enough time to check all the cases her. The solution $a=b=1$ (known to us from Catalan) does show here because with $b=1$ we have $1+1=2$ 
$1+\omega=-\omega^2$ and $1+\omega^2=-\omega$ are both units of $O$.
