If $(A,\mathfrak{m},k)$ is a commutative local ring, $M,N$ are two free modules, and you are given a map $M\to N$ such that the induced map $M\otimes k\to N\otimes k$ is an isomorphism, then was $M\to N$ was an isomorphism to begin with? In the case that $M,N$ are of finite rank there are a number of ways to prove this, one can write the map via a matrix and conclude that the matrix is invertible as the determinant is invertible, or one can use Nakayama's lemma to prove that the cokernel is $0$, then use flatness of $N$ and Nakayama again to show that the kernel is $0$. I attempted to prove this result when we assume that $M,N$ are now free modules of infinite rank, but could not succeed. My personal belief after working on various attempts to solve the question is that the result is not true, but also fail to produce an example. I can however show that one cannot check surjectivity mod $\mathfrak{m}$ in the absence of the assumption that the map is injective mod $\mathfrak{m}$, here is an example that a friend and I produced.
Let $A = \left(k[x_1,x_2,\dots]/(x_{i+1}^2 - x_i)\right)_\mathfrak{m}$ where $\mathfrak{m} = (x_1,x_2,\dots)$ and $k$ is a countable field. One sees that $\mathfrak{m}^2 = \mathfrak{m}$, but is non-zero. Let $A^\omega\to \mathfrak{m}$ be the map sending basis elements $e_i$ to $x_i$, this is a surjection, and then let $A^\omega\to A^\omega$ be a surjection onto the kernel of the map defined above; it is clear that the kernel is countable so such a surjection exists. As $\mathfrak{m}=\mathfrak{m}^2$ we see that the cokernel of $A^\omega\to A^\omega$ vanishes mod $\mathfrak{m}$ which is to say that the map becomes surjective after tensoring with $k$, but the map $A^\omega\to A^\omega$ is not surjective.
A counterexample was brought to my attention, which for the sake of anyone who may have the same question, I will detail here.
Let $k[x]_{(x)}$ be the ring, and let $F$ be a free module of countable rank. Define a map $F\to F$ which as a matrix is represented by $1$ along the diagonal, and $x$ below the diagonal, with $0$ everywhere else. It is clear that this is the identity mod $(x)$. Evaluating this on a tuple $(a_1,\dots,a_n,0,\dots)$ with $a_n$ the final non-zero coordinate, then the image of this under our map will have a non-zero element in the $n+1$-th coordinate which lies in $(x)$, so that the map cannot be surjective. This example generalizes to any local ring $(A,\mathfrak{m})$ such that there exists a non-zero divisor $x\in \mathfrak{m}$, so that in the Noetherian case we see that the result in my original question either fails, or our ring has depth $0$ as its maximal ideal consists of zero-divisors.