= Solution
Suppose first that $x^p+y^p=z^p$ with $x,y,z\in\mathbb Z_p^\times$. Divide by $y^p$ and put $u=x/y$, $w=z/y$. Reduction modulo $p$ gives $w\equiv u+1\pmod p$. Choose an integer $a\equiv u\pmod p$; then $p\nmid a(a+1)$. A unit's $p$th power modulo $p^2$ depends only on its residue modulo $p$, because $(a+pt)^p\equiv a^p\pmod{p^2}$. Therefore
$$
(a+1)^p\equiv w^p=u^p+1\equiv a^p+1\pmod{p^2}.
$$
Conversely, suppose such an integer $a$ exists. The congruence says
$$
\frac{a^p+1}{(a+1)^p}\in1+p^2\mathbb Z_p.
$$
For odd $p$, the $p$th-power map sends $1+p\mathbb Z_p$ onto $1+p^2\mathbb Z_p$: under the <p-adic logarithm> it becomes multiplication by $p$. Hence the displayed ratio is $t^p$ for some $t\in1+p\mathbb Z_p$. Taking
$$
x=a,
\qquad y=1,
\qquad z=(a+1)t
$$
produces the required solution in $\mathbb Z_p^\times$.
Solved by gpt-5.6-sol high.
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