So I need to find MME for $p$ when $X\sim \operatorname{Bi}(n,p)$. $$\operatorname{E}[X] = np$$ $$\operatorname{Var}(X) = np(1-p)$$ I first divided $\operatorname{Var}(X)$ and $\operatorname{E}[X]$ to eliminate $n$. $$\frac{\operatorname{Var}(X)}{\operatorname{E}[X]} = 1-p$$
To solve: $\operatorname{E}[X] = \bar{x}$ and $\operatorname{Var}(X) = \frac{1}{n}\sum x_i^2 - \frac{1}{n}\bar{x}^2$.
But now I am currently stuck at:
$$p=1-\left(\frac{\sum x_i^2-\bar{x}^2}{n\bar{x}}\right).$$
How can I eliminate $n$?
You must be careful to define what constitutes your sample.
For instance, suppose you are observing a sequence of independent and identically distributed (iid) binomial random variables, say $X_1, X_2, X_3, \ldots$, each of which has the same distribution $$X_i \sim \operatorname{Binomial}(n, p).$$ Then the sample size does not necessarily equal the parameter $n$, since you could potentially observe many more data; e.g., your sample could be $X_1, X_2, \ldots, X_m$. Then the sample mean and (bias-uncorrected) sample variance $$\bar X = \frac{1}{\color{red}{m}} \sum_{i=1}^{\color{red}{m}} X_i, \quad \varsigma^2 = \frac{1}{\color{red}{m}} \sum_{i=1}^{\color{red}{m}} \left( X_i - \bar X\right)^2 \tag{1}$$ are calculated for that sample size $m$, and a method of moments estimator for $p$ would be $$\tilde p = 1 - \frac{\varsigma^2}{\bar X} = 1 - \frac{\sum_{i=1}^m (X_i - \bar X)^2}{\sum_{i=1}^m X_i}. \tag{2}$$ There is no $n$ in this expression. Note that we must have $m > 1$ for $(2)$ to be of any value, since the sample variance is $0$ if $m = 1$.
On the other hand, if what you are observing is a sequence of $n$ iid Bernoulli random variables, say $B_1, B_2, \ldots, B_n$, with $$B_i \sim \operatorname{Bernoulli}(p),$$ then in effect the statistic $$X = \sum_{i=1}^n B_i \sim \operatorname{Binomial}(n,p)$$ is a single binomial variable. The method of moments estimator must now be calculated for the Bernoulli sample; i.e.
$$\operatorname{E}[B_i] = p, \quad \operatorname{Var}[B_i] = p(1-p),$$
hence you could write
$$\tilde p = 1 - \frac{\sum_{i=1}^n (B_i - \bar B)^2}{\sum_{i=1}^n B_i}. \tag{3}$$ But this is not necessary: you already have with the first moment
$$\tilde p = \bar B$$ and there is no need to use the sample variance.