Edit Correct a typo and make things more precise.
We want to show that for any $N$ big enough and any $a\in \mathbb Z_p$, there exists a solution with $y=p^Na$ and $x\equiv 1 \mod p$.
If $n$ is prime to $p$, then for any $b\in\mathbb Z_p$, the equation $(1+pw)^n=1+pb$ has a solution with $w\in \mathbb Z_p$. Or, if you prefer, $1+pb$ has a $n$th root in $\mathbb Z_p$, congruent to $1$ mod $p$.
If $n=p$, then for any $b\in\mathbb Z_p$, the equation $(1+pw)^n=1+p^2b$ has a solution with $w\in \mathbb Z_p$.
Both statements are proved by developping the left-hand side, simplifying by $1$ and dividing by $p$ (resp. $p^2$) to get an equation $ F(w)=0$ with $F(w)\in \mathbb Z_p[w]$ (not monic) and $F(w)\equiv w+b \mod p$· Then apply Hensel Lemma which states that there is a solution. Moreover, inspecting the form of $F(w)$, we see that any solution satisfies $v_p(w)=v_p(b)$.
For a general $n$, write $n=mp^r$ with $m$ prime to $p$. Using repeatly (1) and (2) above, we see that for any fixed $a\in\mathbb Z_p$, the equation $(1+pw)^n=1+p^{2r+1}a$ has a solution with $w\in\mathbb Z_p$.
Now back to your equation. Fix any $a\in \mathbb Z_p$ and any $N\ge 2$. We are looking for solutions of the form $y=p^Na$ and $x=1+pw$. Then your equation becomes: $ (1+pw)^n=1-p^{nN}a^n$ (I forgot the $n$th power in the RHS) with $w\in \mathbb Z_p$. As $nN\ge 2r+1$, this equation has a solution. By varying $N$ or $a$, we get infinitely many solutions for $x^n+y^n=1$ with $x=1+pw$ and $y=p^Na$.