$Q(x)$ is a polynomial, so $Q(x)$ is as well.

We now establish a lemma giving an alternate characterization of nice polynomials.

Lemma 2. If $P(x)$ is a polynomial with constant term 1, then $P(x)$ is nice if and only if each nonzero term in the power series expansion of $P(x)/(1-x)$ has coefficient 1.

Proof. Suppose that $P(x) = a_0 + a_1x + \cdots$. Notice that the power series of $P(x)$ has coefficients
$$\frac{P(x)}{1-x} = b_0 + b_1x + b_2x^2 + \cdots = a_0 + (a_0 + a_1)x + (a_0 + a_1 + a_2)x^2 + \cdots$$
given by the partial sums of the coefficients of $P(x)$. (In particular, this means that the coefficients of the power series of $\frac{P(x)}{1-x}$ are eventually constant.) Because $P(0) = 1$, we have $b_0 = a_0 = 1$.

Now, if $P(x)$ is nice, then because the nonzero coefficients of $P(x)$ alternate between 1 and $-1$, the partial sums of the coefficients of $P(x)$ take value either 0 or 1. This means exactly that $b_i \in \{0, 1\}$, hence all non-zero coefficients of the power series for $\frac{P(x)}{1-x}$ are equal to 1, as desired.

On the other hand, if $b_i \in \{0, 1\}$, we see that $a_i = b_i - b_{i-1} \in \{-1, 0, 1\}$. Explicitly, this means that
$$a_i = \begin{cases} 1 & b_i > b_{i-1} \\ 0 & b_i = b_{i-1} \\ -1 & b_i < b_{i-1} \end{cases}.$$
Because $b_i$ takes at most two values, among $i$ for which $b_i \neq b_{i-1}$, the first and last cases alternate, which implies exactly that $P(x)$ is nice. This completes the proof of the lemma. $\square$

We now turn to the problem proper. Because $P(0) = 1$, it follows that $Q(0) = 1$. Thus, by Lemma 2, it suffices to show that all nonzero terms in the power series for
$$\frac{Q(x)}{1-x} = \frac{P(x^n)}{1-x^n} \cdot \frac{1-x^{mn}}{1-x^m}$$
have coefficient 1. Again by Lemma 2, all nonzero terms in the power series of $P(x)/(1-x)$ have coefficient 1, so the same is true for $\frac{P(x^n)}{1-x^n}$. Further, all the nonzero terms of the power series expansion of $\frac{P(x^n)}{1-x^n}$ have exponents congruent to 0 modulo $n$. Now, because $m$ and $n$ are relatively prime, $\frac{1-x^{mn}}{1-x^m}$ is a polynomial whose nonzero coefficients are equal to 1 and whose nonzero terms have exponents with distinct residues modulo $n$. Therefore, each nonzero term in the power series expansion of $\frac{Q(x)}{1-x}$ may be expressed in a unique way as the product of a nonzero term in the power series of $\frac{P(x^n)}{1-x^n}$ and a nonzero term in $\frac{1-x^{mn}}{1-x^m}$, hence each such term has coefficient 1. So $Q(x)$ is nice by Lemma 2.