This is a special case $\,f(x) = x^k-a\, $ of the result below, which shows that if $\,m,n\in\Bbb Z\,$ are coprime and $\,f\in\Bbb Z[x]\,$ is a polynomial with integer coefs, then the roots of $\,f\bmod mn$ correspond to CRT-combining the roots of $f\bmod m\,$ and $\,f\bmod n.\,$ In particular this yields the sought existence claim:
$$ f\,\ \text{is solvable $\!\bmod mn\iff f\,$ is solvable $\!\bmod m\,$ & $\!\bmod n$}\qquad$$
Suppose that $\,f(x)\,$ is a polynomial with integer coefs and $\,m,n\,$ are coprime integers. By CRT, solving $\,f(x)\equiv 0\pmod{\!mn}\,$ is equivalent to solving $\,f(x)\equiv 0\,$ mod $\,m\,$ and mod $\,n,\,$ and each CRT combination of a root $\,r_i\,$ mod $\,m\,$ with a root $\,s_j\,$ mod $\,n\,$ corresponds to a unique root $\,t_{ij}\bmod mn\,$ i.e.
$$\begin{eqnarray} f(x)\equiv 0\!\!\!\pmod{\!mn}&\overset{\rm CRT}\iff& \begin{array}{}f(x)\equiv 0\pmod{\! m}\\f(x)\equiv 0\pmod{\! n}\end{array} \\
&\iff& \begin{array}{}x\equiv r_1,\ldots,r_k\pmod {\!m}\phantom{I^{I^{I^I}}}\\x\equiv s_1,\ldots,s_\ell\pmod{\! n}\end{array}\\
&\iff& \left\{ \begin{array}{}x\equiv r_i\pmod{\! m}\\x\equiv s_j\pmod {\!n}\end{array} \right\}_{\begin{array}{}1\le i\le k\\ 1\le j\le\ell\end{array}}^{\phantom{I^{I^{I^I}}}}\\
&\overset{\rm CRT}\iff& \left\{ x\equiv t_{i j}\!\!\pmod{\!mn} \right\}_{\begin{array}{}1\le i\le k\\ 1\le j\le\ell\end{array}}\\
\end{eqnarray}\qquad\qquad$$
You can find many concrete worked examples of this isomorpism in prior posts, e.g. below
\pmod{x}
to generate $\pmod{x}$ (note the spacing). $\endgroup$