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 $\phi$ <i>(added: see below for a variant where I justify this claim)</i> . 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).

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<b>Added:</b> here's a little variant in which I can provide a short argument for the statement on the order: define the <i>wreathed Coxeter group</i>

$$H=\langle t,x\mid x^2,s_{\phi(n)}^{n!}:n\ge 1\rangle,\quad \text{where}\;s_m=x_0x_m,\;x_m=t^mxt^{-m}.$$

Indeed we immediately see that $H$ is a semidirect product $\mathbf{Z}\ltimes W$, where $W$ is the Coxeter group with generators $x_m$, $m\in\mathbf{Z}$, and Coxeter relations $(x_{m+\phi(n)}x_m))^{n!}=1$, where the cyclic group acts by shifting the $x_m$. And it's well-known that in a Coxeter group, the prescribed orders are the genuine orders [we just have to check that we injectively prescribe order, which amounts of the injectivity of $(m,n)\mapsto (m+\phi(n),m)$, which itself follows from injectivity of $\phi$].

By the way, it follows that the same holds if we remove the relator $x^2=1$ in the above presentation.