I can think of some really contrived or annoying arguments, but my professor gave a simple explanation for this (and general questions like this, such as when there exists a $q^{th}$ root of unity mod $p$), but I can't remember it.
If anyone could help me out that'd be great!
On
Why don't you try a bit of modular arithmetic to get a neat general answer (perhaps your professor did this) ? For any integer $n={p_1}^{r_1}...{p_s}^{r_s}$ (prime factorization), the Chinese Remainder theorem gives an isomorphism of rings $\mathbf Z /n \cong \mathbf Z /{p_1}^{r_1} \times ...\times \mathbf Z /{p_s}^{r_s}$, which gives in turn an isomorphism $(\mathbf Z /n)^* \cong (\mathbf Z /{p_1}^{r_1})^* \times ...\times (\mathbf Z /{p_s}^{r_s})^*$ between the multiplicative groups of invertible elements. For a given prime $p$, it is known that $(\mathbf Z /{p}^{r})^*$ is cyclic of order $p({p}^{r}-1)$ if $p$ is odd, $(\mathbf Z /{2}^{r})^* \cong C_2 \times C_{r-2}$ if $p=2$, where the cyclic groups $C_2, C_{r-2}$ are generated resp. by the classes of $1$ and $5$. Now, writing any class of $\mathbf Z /n$ under the form $x=(x_1,...,x_s)$, we have $x^q=1$ iff $x_j^q=1$ for $j=1,...,r$. Lagrange theorem tells you that if $p$ is odd, $(\mathbf Z /{p}^{r})^*$ contains a (unique) subgroup of order $q$ iff $q \mid p({p}^{r}-1)$. If $p=2$, the subgroups of $(\mathbf Z /{2}^{r})^*$ are obtained by composing those of $C_2$ and $C_{r-2}$. In your case here, $x^4=1, p=2, r=3, q=4$, so $(\mathbf Z /8)^* \cong C_2 \times C_2$ and the only possibilities are $<x>= <2>$, or $<5>$, or $<2> \times <5>$ .
Take any odd integer $x$. Then $$ 8 \mid x^2-1=(x+1)(x-1) $$ (both factors are even, and one of them divisible by $4$). In particular, $8\mid (x^2)^2-1$.