In this paper, we consider the problem of finding a cycle of length $2k$ (a $C_{2k}$) in an undirected graph $G$ with $n$ nodes and $m$ edges for constant $k\ge2$. A classic result by Bondy and Simonovits [J.Comb.Th.'74] implies that if $m \ge100k n^{1+1/k}$, then $G$ contains a $C_{2k}$, further implying that one needs to consider only graphs with $m = O(n^{1+1/k})$. Previously the best known algorithms were an $O(n^2)$ algorithm due to Yuster and Zwick [J.Disc.Math'97] as well as a $O(m^{2-(1+\lceil k/2\rceil^{-1})/(k+1)})$ algorithm by Alon et al. [Algorithmica'97]. We present an algorithm that uses $O(m^{2k/(k+1)})$ time and finds a $C_{2k}$ if one exists. This bound is $O(n^2)$ exactly when $m=Θ(n^{1+1/k})$. For $4$-cycles our new bound coincides with Alon et al., while for every $k>2$ our bound yields a polynomial improvement in $m$. Yuster and Zwick noted that it is "plausible to conjecture that $O(n^2)$ is the best possible bound in terms of $n$". We show "conditional optimality": if this hypothesis holds then our $O(m^{2k/(k+1)})$ algorithm is tight as well. Furthermore, a folklore reduction implies that no combinatorial algorithm can determine if a graph contains a $6$-cycle in time $O(m^{3/2-ε})$ for any $ε>0$ under the widely believed combinatorial BMM conjecture. Coupled with our main result, this gives tight bounds for finding $6$-cycles combinatorially and also separates the complexity of finding $4$- and $6$-cycles giving evidence that the exponent of $m$ in the running time should indeed increase with $k$. The key ingredient in our algorithm is a new notion of capped $k$-walks, which are walks of length $k$ that visit only nodes according to a fixed ordering. Our main technical contribution is an involved analysis proving several properties of such walks which may be of independent interest.