Can we find two positive integers $n$ and $m$ ($n,m>1$) such that $n^\pi = m$?

(Turning comments into an answer, as requested)

This follows from Schanuel's conjecture but it's probably hard to prove unconditionally.

Apply Schanuel to $2\pi i,\log n,\log m$. The last two numbers are linearly independent over $\mathbb Q$ because of your hypothesis and the fact that $\pi$ is irrational. Then all three numbers are linearly independent over $\mathbb Q$ since $2\pi i$ is not real. Finally, the exponentials of all three numbers are rational, so Schanuel implies that the three numbers are algebraically independent over $\mathbb Q$ which is stronger than what you want.

Unconditionally, this can essentially happen at most once. It is a consequence of the Five Exponentials Theorem, which itself follows from a combination of the classical Six Exponentials Theorem and Baker's Theorem on linear forms in logarithms.

Five Exponentials Theorem: (See Waldschmidt, Diophantine Approximation on Linear Alebgraic Groups, 2000, Section 11.3.3)

Suppose that $\lambda_0, \lambda_1, \lambda_2 \in \mathbb{C}^{\times}$ are chosen so that $e^{\lambda_i} \in \overline{\mathbb{Q}}$ for each $i = 0, 1, 2$. Suppose that $\lambda_1$ and $\lambda_2$ are linearly independent over $\mathbb{Q}$. Then for any non-zero algebraic number $\beta$, at least one of the two numbers $$ e^{\beta \lambda_0 \lambda_1}, \quad e^{\beta \lambda_0\lambda_2} $$ is transcendental.

To apply to the present question, for positive integers $n_1$ and $n_2$ take $$ \lambda_0 = \pi i, \quad \lambda_1 = \log n_1, \quad \lambda_2 = \log n_2, \quad \beta = -i, $$ and so $$ e^{\beta \lambda_0 \lambda_1} = n_1^{\pi}, \quad e^{\beta \lambda_0 \lambda_2} = n_2^{\pi}. $$ By the Five Exponentials Theorem, if both $n_1^{\pi}$ and $n_2^{\pi}$ are algebraic, then $n_1$ and $n_2$ must be multiplicatively dependent. Thus up to multiplicative dependence there is at most one way for a solution to be found.