exist $x$ such that $x^k \equiv m$ mod $(p_1\cdot p_2) \Leftrightarrow $ exists $x_1,x_2$ : $x_1^k\equiv m(p_1)$ and $x_2^k\equiv m(p_2)$

exist $ x$ such that $ x^k \equiv m$ mod $ (p_1\cdot p_2) \Leftrightarrow $ exists $ x_1,x_2$ : $ x_1^k\equiv m(p_1)$ and $ x_2^k\equiv m(p_2)$

A first approach I took, was to use $ y\equiv m(p_1) , y\equiv m(p_2) \Leftrightarrow y\equiv m(p_1\cdot p_2)$ , then by assigning $ x^k=y$ the problem comes to find whether $ y$ has a $ k$ -order root in $ U(\mathbb{Z}_{p_1\cdot p_2})$ . How ever it doesn’t seem to simplify the problem.

A second approach I took was to use the fact which derived from CRT , that $ U(\mathbb{Z}_{p_1 \cdot p_2}) \cong U(\mathbb{Z}_{p_1}) \times U(\mathbb{Z}_{p_2}) $ , In $ U(z_{p_i})$ which are cyclic groups, there is a solution for $ x^k \equiv m(p_i) \Leftrightarrow m^{\frac{p_1-1}{gcd(k,p_1)}}=1 (p_i)$ . So assuming $ gcd(k,p_1) = 1$ there are solutions for the equations $ x_1,x_2$ . But I am struggling to show that $ \pi^{-1}(x_1,x_2)$ (when $ \pi$ is the isomorphism from CRT), is a solution for $ x^k \equiv m (p_1p_2)$ .

So in case my second approach is correct, I would be glad for some help with showing $ \pi^{-1}(x_1,x_2)$ is a solution, and also in case $ x$ is a solution mod $ (p_1p_2)$ then $ \pi_1(x) ,\pi_2(x)$ are solutions mod $ p_1$ , $ p_2$ respectively.

Also other approaches or ideas would be appreciated.

related question:

If $ x \equiv a \pmod {p_1}$ and $ x\equiv a \pmod{p_2}$ , then is it true that $ x\equiv a \pmod{p_1p_2} ?$