If $x+i=(a+ib)^p$, then $x-i=(a-ib)^p$, $2i=(a+ib)^p-(a-ib)^p=2ib$(smth), $b=\pm 1$, $1=\pm 1\mp \binom{p}2a^2\pm \binom{p}4 a^4+\dots$. Considering this modulo $a^2$ we see that signs start from plus, that is, $\binom{p}2a^2\pm \binom{p}4 a^4+\dots=0$. Next, $\binom{p}{2k}=\binom{p}2\binom{p-2}{2k-2}/\binom{2k}2$ and dividing by $\binom{p}2a^2$ we get
$$
1=\sum_{k\geqslant 2} \pm \binom{p-2}{2k-2}\frac{a^{2k-2}}{k(2k-1)}.
$$
If $a$ has some prime divisor $q$, then RHS has positive $q$-adic valuation, since $k(2k-1)$ can not be divisible by $q^{2k-2}$. If $a=\pm 1$, then we know what is $(a+i)^p$ and it is not $x+i$.

This is possibly not what you were asking for, but at least it solves this case of Catalan conjecture very quickly.