Yes. Let $F$ be a nonabelian f.g. free group, then $F\times F$ has this property. Indeed, for each normal subgroup $N$ of $F$, denote $H_N$ the subgroup $\{(x,y), xN=yN\}$ of $F\times F$ (this is a fibre product $F\times_{F/N} F$). Then $H_N$ contains the normal subgroup $N\times N$, and the quotient $H_N/(N\times N)$ is isomorphic to $F/N$. This normal subgroup can be intrinsically defined in $H_N$, because it is generated by $N\cup\{1\}\cup \{1\}\times N$, which is the set of elements of $H_N$ whose centralizer in $H_N$ is nonabelian. Hence the isomorphism class of $H_N$ determines the isomorphism class of $F/N$. Moreover, if $F/N$ is simple then $H_N$ is maximal in $F\times F$. Since there are uncountably many non-isomorphic 2-generated simple groups, we deduce that $F\times F$ has uncountably many non-isomorphic maximal subgroups.

Edit: here's a reference for the fact that there exists uncountably many non-isomorphic 2-generated simple groups:

Camm, Ruth.
Simple free products.
J. London Math. Soc. 28, (1953). 66-76.

[MR review behind paywall: (...) we have a continuum of non-isomorphic simple groups $G(\rho,\sigma,\tau)$. Each of these is a free product, with an amalgamated subgroup, of two free groups, and so is, in particular, locally infinite. Moreover, each can be generated by two elements.]