Inspired by McCool's paper given in Benjamin's answer, here's an explicit example that is finitely generated but not finitely presented:
let $\phi$ be an injective recursive function from positive integers to themselves, whose image is not recursive. Consider the group with recursive presentation $$G=\langle t,x\mid r_{\phi(n)}^{n!}:n\ge 1\rangle,\quad \text{where}\;r_m=[t^mxt^{-m},x].$$ It is not hard to check that $r_{\phi(n)}$ has order $n!$ and $r_m$ has infinite order if $m$ is not in the range of $\phi$. In particular, the infinite order problem (checking if an element has infinite order) is not solvable.
However this group has a solvable word problem. The idea is that given a word of length $n$, it is trivial in $G$ if and only if it is trivial in the partial presentation with only relators $r_{\phi(k)}^{k!}$ for $k\le n$, and word problem in these groups are (I think) simultaneously solvable although I haven't checked details.
Actually McCool says it's enough to embed such a group into a group with solvable word problem, no need to care that the image is recursive. And indeed that's enough (clearly solvability of the infinite order problem passes to finitely generated subgroups).