So I know I'm looking to find a function $\delta(X)$ so that $E\left[\delta(X)\right] = p^2$.
Let me venture a guess that $\delta = \hat{p}^2 = \frac{x^2}{n^2}$. Then:
$$E\left[\hat{p}^2 \right] = E\left[ \frac{x^2}{n^2} \right] = \frac{1}{n^2} E\left[x^2\right] = \frac{1}{n^2}\left[Var(X) + E[X]^2 \right] = \frac{1}{n^2}\left(\frac{p(1-p)}{n} + n^2p^2 \right)$$.
So is it ok to say that an unbiased estimator is: $$ \frac{1}{n^2}\left(\frac{p(1-p)}{n} + n^2p^2 \right)-p^2?$$
I'm not feeling confident in this answer since I haven't really taken a statistics class before, so I'm very new to this.
Your estimator should be of the form $\delta(X)$, as you say -- your expression is not, as written it actually assumes you already know $p$.
Outline: the variance of the binomial will be $\operatorname{Var}(X) = \mathbb{E}[X^2] - \mathbb{E}[X]^2 = np(1-p) = np - np^2$.
Since you want $p^2$, you may rewrite this as $$ p^2 = \frac{1}{n}\left( np - \operatorname{Var}(X) \right) = \frac{1}{n}\left( \mathbb{E}[X] - \operatorname{Var}(X) \right). $$ Do you know any unbiased estimators for respectively $\mathbb{E}[X]$ and $\operatorname{Var}(X)$? If so, can you combine them to get one for $p^2$?