Every point has a fundamental system of neighborhoods given by $\{x+p^n\Bbb Z_p\}_{n\in\Bbb N}$ which are compact.
In essence this is just the fact that $p^n\Bbb Z_p$ is a fundamental system of compact neighborhoods of $0$: since we are in a vector space--really a topological group is enough, but not everyone is familiar with structures of that generality--we can translate these sets anywhere to form a compact (and open) neighborhood of any point.
If you have any trouble seeing this recall that open balls generate the topology and all open balls are of the form $x+p^n\Bbb Z_p$ for some $n\in\Bbb Z$.
I don't offhand see any way to use the binomial expansion to solve your question. Instead, here's an alternate approach. Because $[x]_p$ has multiplicative order $3$, then $x \not\equiv 1 \pmod{p}$ and
$$x^3 \equiv 1 \pmod{p} \implies x^3 - 1 \equiv 0 \pmod{p} \tag{1}\label{eq1A}$$
Note the lifting-the-exponent lemma states that, for odd primes $p$ where $p \not\mid x$ and $p \not\mid y$, then
- If $p \mid x - y$, $\; \nu _{p}(x^{n}-y^{n}) = \nu _{p}(x - y) + \nu _{p}(n)$.
- If $n$ is odd and $p \mid x + y$, $\; \nu _{p}(x^{n} + y^{n}) = \nu _{p}(x + y) + \nu _{p}(n)$.
Thus, using \eqref{eq1A} and the first option above gives
$$\nu_p((x^{3})^{p} - 1) = \nu_p(x^3 - 1) + \nu_p(p) \ge 2 \tag{2}\label{eq2A}$$
We therefore have
$$x^{3p} - 1 \equiv 0 \pmod{p^2} \tag{3}\label{eq3A}$$
Using Fermat's little theorem gives $x^{p} - 1 \equiv x - 1 \not\equiv 0 \pmod{p}$. Since $x^{3p} - 1 = (x^p - 1)(x^{2p} + x^{p} + 1)$, we thus get from \eqref{eq3A} that
$$x^{2p} + x^{p} + 1 \equiv 0 \pmod{p^2} \implies x^{p} + 1 \equiv -x^{2p} \pmod{p^2} \tag{4}\label{eq4A}$$
Since $x^3 = (x - 1)(x^2 + x + 1)$ and $x \not\equiv 1 \pmod{p}$, we thus get from \eqref{eq1A} that
$$x^2 + x + 1 \equiv 0 \pmod{p} \tag{5}\label{eq5A}$$
Next, using the second option of the lifting-the-exponent lemma statement mentioned earlier gives
$$\nu_p((x^{2})^{p} + (x + 1)^p) = \nu_p(x^2 + x + 1) + \nu_p(p) \ge 2 \tag{6}\label{eq6A}$$
From this and using \eqref{eq4A}, we get
$$\begin{equation}\begin{aligned}
x^{2p} + (x + 1)^p & \equiv 0 \pmod{p^2} \\
(x + 1)^p & \equiv -x^{2p} \pmod{p^2} \\
(x + 1)^p & \equiv x^p + 1 \pmod{p^2}
\end{aligned}\end{equation}\tag{7}\label{eq7A}$$
Best Answer
Lemma: Let $p \neq 2$ be a prime, $a,b$ and $c$ be pairwise coprime integers with $abc$ square-free and $p|a$, and $Q: ax^2 + by^2 + cz^2 = 0$ a quadratic form. Then there is a nontrivial solution to $Q$ over $\mathbb{Q}_p$ if and only if $-b/c$ is a square mod $p$.
Proof: Suppose a solution exists. Then by scaling, we may assume that $y$ and $z$ lie in $\mathbb{Z}_p$, and indeed that they lie in $\mathbb{Z}_p^\times$. (Suppose without loss of generality that $y \in p\mathbb{Z}_p;$ then as $p|a$, $p|y$ and $p\nmid c$, we must have $p|z,$ and hence $p|x$ by considering the parity of the exponent of $p$ in the sum. So we can divide our trio $(x,y,z)$ by $p$.)
Now consider the equation mod $p$. This becomes equivalent to $(y/z)^2 \equiv -b/c$ mod $p,$ where we can divide by $c$ and $z^2$ as they have invertible image in $\mathbb{F}_p$. So $y/z$ gives the corresponding element of $\mathbb{F}_p$ that squares to $-b/c$.
Conversely, suppose that we have $Y^2 \equiv -b/c$ mod $p$. Then in particular, $p\nmid Y$ and the triple ($x,y,z) = (0,Y,1)$ gives a solution mod $p$. Using Hensel's lemma (and using that $p\neq 2$) we see that this lifts to a solution in $\mathbb{Q}_p$, as required.
Reference: At which p-adic fields does the equation have no solution?