Annoying problem. Use $(p+1)(2p+1)=(q+1)(q-1)^2$ to get $p(2p+3) = q(q^2-q-1)$. Check that $p\ne q$ (otherwise $q\in\{1,4\}$), so $p\mid q^2-q-1$ and $q\mid 2p+3$. Set $p=(qk-3)/2$ and arrive at $qk-3\mid 2q^2-2q-2$. Note that $q\equiv 3/k \pmod{qk-3}$, so $2q^2-2q-2\equiv 6q/k -2q-2\pmod{qk-3}$, yielding $qk-3\mid 6q-2qk-2k$. Since we also have $qk-3\mid 2qk-6$, we obtain $qk-3\mid 6q-2k-6$. Now if $3q-3=k$, then $3\mid p$, which can easily seen to yield no solutions. So, $|6q-2k-6|>0$. Note that if $2k+6>6q$ then $qk-3\le 2k+6-6q$ yields a contradiction. So, $6q-2k-6>0$, from some simple inequalities we get $k\in\{1,\dots,5\}$. Some algebra gives $(p,q)=(31,13)$ as the only solution, so $a=2016$. We get $2015+ 2^b+3^c=3d!$. Now I show the only solutions are $(a,b,c,d)=(2016,6,4,6)$. To that end, assume the contrary. Using modulo 3, we find $b$ even. Next, using modulo 4, we find $c$ even, too. Furthermore, $d!>2015/3$, so $d\ge 6$. For $d=6$, we obtain the quadruple above, so suppose $d\ge 7$. Since $c$ is even, $c\ge 2$ so by modulo 9, we find $2^b\equiv 1\pmod{9}$. In particular, $6\mid b$. Now, using $2^b\equiv 1\pmod{7}$ by Fermat, and the fact $d\ge 7$, we get $3^c\equiv 0\pmod{7}$, which is absurd.