We will prove that the answer is no. Note that

$$m^{3}=9 n^{4}+170 n^{2}+289=\left(9 n^{2}+17\right)\left(n^{2}+17\right)$$

If $n$ is odd then $m$ is even, therefore $8 \mid m^{3}$. However,

$$9 n^{4}+170 n^{2}+289 \equiv 9+170+289 \equiv 4(\bmod 8)$$

which leads to a contradiction. If $n$ is a multiple of 17 then so is $m$ and hence 289 is a multiple of $17^{3}$, which is absurd. For $n$ even and not multiple of 17, since

$$\operatorname{gcd}\left(9 n^{2}+17, n^{2}+17\right) \mid 9\left(n^{2}+17\right)-\left(9 n^{2}+17\right)=2^{3} \cdot 17$$

this gcd must be 1. Therefore $n^{2}+17=a^{3}$ for an odd $a$, so

$$n^{2}+25=(a+2)\left(a^{2}-2 a+4\right)$$

For $a \equiv 1(\bmod 4)$ we have $a+2 \equiv 3(\bmod 4)$, while for $a \equiv 3(\bmod 4)$ we have $a^{2}-2 a+4 \equiv 3$ $(\bmod 4)$. Thus $(a+2)\left(a^{2}-2 a+4\right)$ has a prime divisor of type $4 \ell+3$. As it divides $n^{2}+25$, it has to divide $n$ and 5, which is absurd.

1. We start with the given equation:
$$m^3 = 9n^4 + 170n^2 + 289$$
We can rewrite the right-hand side as a product of two polynomials:
$$m^3 = (9n^2 + 17)(n^2 + 17)$$

2. Next, we consider the case where $17$ divides $n$. If $17 \mid n$, then $n = 17k$ for some integer $k$. Substituting $n = 17k$ into the equation, we get:
$$m^3 = 9(17k)^4 + 170(17k)^2 + 289$$
Simplifying, we have:
$$m^3 = 9 \cdot 17^4 k^4 + 170 \cdot 17^2 k^2 + 289$$
Since $17^3 \mid 9 \cdot 17^4 k^4$ and $17^3 \mid 170 \cdot 17^2 k^2$, it follows that $17^3 \mid m^3$. Therefore, $17 \mid m$. Let $m = 17p$ for some integer $p$. Substituting $m = 17p$ into the equation, we get:
$$(17p)^3 = 9 \cdot 17^4 k^4 + 170 \cdot 17^2 k^2 + 289$$
Simplifying, we have:
$$17^3 p^3 = 9 \cdot 17^4 k^4 + 170 \cdot 17^2 k^2 + 289$$
Dividing both sides by $17^3$, we get:
$$p^3 = 9 \cdot 17 k^4 + 170 k^2 + \frac{289}{17^3}$$
Since $\frac{289}{17^3}$ is not an integer, this leads to a contradiction. Hence, $17$ does not divide $n$.

3. Now, we consider the case where $17$ and $n$ are coprime. Let $\text{gcd}(9n^2 + 17, n^2 + 17) = d$. It follows that $d \mid (9n^2 + 17) - 9(n^2 + 17)$:
$$d \mid (9n^2 + 17) - 9(n^2 + 17) = 9n^2 + 17 - 9n^2 - 153 = -136$$
Therefore, $d \mid 136$. The divisors of 136 are $1, 2, 4, 8, 17, 34, 68, 136$.

4. If $d = 1$, assume we have $9n^2 + 17 = a^3$ and $n^2 + 17 = b^3$ for positive integers $a$ and $b$. Considering modulo 7:
$$9n^2 + 17 \equiv 2n^2 + 3 \pmod{7}$$
$$n^2 + 17 \equiv n^2 + 3 \pmod{7}$$
Since $a^3 \equiv b^3 \pmod{7}$, we need $2n^2 + 3 \equiv n^2 + 3 \pmod{7}$, which implies $n^2 \equiv 0 \pmod{7}$. Thus, $7 \mid n$, contradicting that $17$ and $n$ are coprime.

5. If $d = 2$, we have $8 \mid (9n^2 + 17)(n^2 + 17)$. Considering modulo 4:
$$9n^2 + 17 \equiv n^2 + 1 \pmod{4}$$
$$n^2 + 17 \equiv n^2 + 1 \pmod{4}$$
Since $8 \mid (n^2 + 1)(n^2 + 1)$, we need $n^2 \equiv 3 \pmod{4}$, which is impossible since $n^2 \equiv 0, 1 \pmod{4}$.

6. If $d = 4$ or $d = 8$, we will get a contradiction by considering modulo 4:
$$9n^2 + 17 \equiv n^2 + 1 \pmod{4}$$
$$n^2 + 17 \equiv n^2 + 1 \pmod{4}$$
Since $4 \mid (n^2 + 1)(n^2 + 1)$, we need $n^2 \equiv 3 \pmod{4}$, which is impossible.

Therefore, there are no positive integers $m$ and $n$ satisfying the equation $m^3 = 9n^4 + 170n^2 + 289$.