If aᵖ ≡ bᵖ (mod p) then aᵖ ≡ bᵖ (mod p²)

Step 1. By Fermat's little theorem $a^p\equiv a$ and $b^p\equiv b\pmod p$. The hypothesis gives $a\equiv b\pmod p$, so $a=b+kp$ for some integer $k$.

Step 2: binomial expansion.
$$a^p=(b+kp)^p=b^p+\binom p1b^{p-1}kp+\sum_{j\ge2}\binom pjb^{p-j}(kp)^j.$$
The second term is $p\cdot b^{p-1}\cdot kp$, a multiple of $p^2$. For $j\ge2$ the factor $(kp)^j$ is a multiple of $p^2$. Hence $a^p\equiv b^p\pmod{p^2}$.

Example. $17\equiv2\pmod5$, and indeed $17^5-2^5=1419825=25\cdot56793$.

Remark. First step of "lifting the exponent": $v_p(a^p-b^p)=v_p(a-b)+1$ when $p\mid a-b$ and $p$ is odd.

Lesson. When two numbers are congruent modulo $p$, write one as the other plus a multiple of $p$; the binomial expansion then shows the $p$-th powers agree one power of $p$ higher.