Is there a general formula for $I(m,n)$?

We can use contour integration to arrive at the general result. Note that

$$\begin{align} \oint_C \frac{z^m}{z^n+1}\,dz&=2\pi i \text{Res}\left(\frac{z^m}{z^n+1}, z=e^{i\pi/n}\right)\\\\ &=-2\pi i \frac{e^{i\pi(m+1)/n}}{n}\tag 1 \end{align}$$

where $C$ is the "pie slice" contour comprised of (i) the real-line segment from $0$ to $R$, where $R>1$, (ii) the circular arc of radius $R$ that begins at $R$ and ends at $Re^{i2\pi/n}$, and $(3)$ the straight line segment from $Re^{i2\pi/n}$ to $0$.

Then, we can write

$$\oint_C \frac{z^m}{z^n+1}\,dz=\int_0^R \frac{x^m}{x^n+1}\,dx+\int_0^{2\pi/2}\frac{R^me^{im\phi}}{R^ne^{in\phi}+1}\,iRe^{i\phi}\,d\phi-\int_0^R \frac{x^me^{i2\pi m/n}}{x^n+1}e^{i2\pi/n}\,dx \tag 2$$

If $n>m+1$, then as $R\to \infty$, the second integral on the right-hand side of $(2)$ vanishes and we find that

$$\bbox[5px,border:2px solid #C0A000]{\int_0^\infty \frac{x^m}{x^n+1}\,dx=2\pi i\frac{e^{i\pi(m+1)/n}}{n(e^{i2\pi(m+1)/n}-1)}=\frac{\pi/n}{\sin(\pi(m+1)/n)}}$$

We shall compute it in two steps. First, perform the substitution $y = x^n$ in order to get

$$I(m,n) = \int \limits _0 ^\infty \frac {y ^{\frac m n}} {1 + y} \frac 1 n y ^{\frac 1 n - 1} \ \Bbb d y = \frac 1 n \int \limits _0 ^\infty \frac {y ^{\frac {m+1} n - 1}} {1 + y} \ \Bbb d y .$$

Now perform the change $t = \frac y {1+y}$, to obtain

$$I(m,n) = \frac 1 n \int \limits _0 ^1 \frac {\left( \frac t {1-t} \right) ^{\frac {m+1} n - 1}} {1 + \frac t {1-t}} \frac 1 {(1-t)^2} \ \Bbb d t = \frac 1 n \int \limits _0 ^1 t^{\frac {m+1} n - 1} (1-t)^{- \frac {m+1} n} \ \Bbb d t = \frac 1 n B \left( \frac {m+1} n, 1 - \frac {m+1} n \right) = \frac 1 n \frac \pi {\sin \pi {\frac {m + 1} n}} .$$

In the above, $B$ is Euler's Beta function and I have used the known formula $B(x, 1-x) = \frac \pi {\sin \pi x}$.