Given your comment about curves I suppose by a one dimensional ideal you mean an ideal such that the ring mod this ideal is one-dimensional.
The answer is this: if you assume your ring to be noetherian yes, if not no.
Let $(A,m)$ be a local ring.
Case 1: $A$ is noetherian. Take the set of prime ideals other than $\mathfrak m$. By the noetherian assumption this set has a (possibly more than one) maximal element. Call that $\mathfrak p$. Then the quotient ring $A/ \mathfrak p$ is a local domain that has exactly two prime ideals, so it has dimension one.
Case 2: $A$ is not necessarily noetherian. Karl Schwede constructs an example in this paper that has a single infinite chain of prime ideals $(0)\subset \mathfrak p_1\subset \mathfrak p_2\subset \dots \subset \mathfrak m\subset A$. He does this to exhibit the scheme $(\mathrm{Spec} A)\setminus\mathfrak m$ as a scheme without a closed point, but it also works here. Since all the prime ideals of $A$ appear in that chain and there are infinitely many "above" any of them except $\mathfrak m$, this ring does not have a prime ideal "of dimension one". In fact taking any prime other than $\mathfrak m$ gives you an infinite dimensional quotient ring. This implies that if the quotient ring by any ideal is finite dimensional, then it has to be 0 dimensional: Take the quotient, if it has finite dimension then the only prime ideal it can have is the image of $\mathfrak m$ and hence its 0-dimensional.
