Completing the squares and setting $X = 6a-1$, $Y = 6b-1$, $Z = 6c-1$ and $W = 6d-1$ we find
$$ X^2 + Y^2 = Z^2 + 1, \quad X^2 - Y^2 = W^2 - 1. $$

Adding these equations we get
$$ 2X^2 = Z^2 + W^2, $$
which we can parametrize by
$$ X = t^2 - 2tu + 2u^2, \quad W = t^2 - 4tu + 2u^2, \quad Z = t^2 - 2u^2. $$
This parametrization yields integral solutions as $t$ and $u$ run
through ${\mathbb Z}$, although perhaps not all of them since these may have a
common divisor.

Neglecting this problem for now we can plug this parametrization into
$$ 2Y^2 = Z^2 - W^2 + 2 $$
and get
$ Y^2 = 4ut^3 - 12u^2t^2 + 8u^3t + 1 $, that is,
$$ (1) \qquad \qquad Y^2 = 4tu(t-u)(t-2u) + 1. $$

** 1. Brute Force **
A direct search for points on this surface yields several solutions;
the solutions with $t < (2-\sqrt{2}\,)u$ give rise to values of
$X$, $Y$, $-Z$, $W$ that are positive, and if $t \equiv 3 \bmod 6$
and $u \equiv \pm 1 \bmod 3$ these values are all $\equiv -1 \bmod 6$,
hence yield positive integral solutions $(a,b,c,d)$ of our problem
The two smallest out of more than a dozen found this way are
$(t,u,a,b,c,d) = (9 , 85, 2167, 1020, 2395, 1912)$ and

$(t,u,a,b,c,d) = (51, 2506, 2051177, 415877, 2092912, 2008575)$.

This suggests that the diophantine problem
has infinitely many solutions in positive integers. Proving this
conjecture seems to be difficult, however.

** 2. Elliptic Surfaces**

Since $Y$ must be odd, we can set
$Y = 2y+1$ and get
$$ y^2 + y = ut^3 - 3u^2t^2 + 2u^3t = tu(t-u)(t-2u). $$
This is an elliptic surface, i.e., an elliptic curve
over the field ${\mathbb Q}(u)$. Obvious rational points are
$(t,y) = (0,0), (0,-1), (u,0), (u,-1), (2u,0), (2u,-1)$.

For transforming this into Weierstrass form, multiply
through by $u^2$ and set $yu = z$, $tu = x$, giving
$$ z^2 + uz = x^3 - 3u^2x^2 + 2u^4x = x(x-u^2)(x-2u^2). $$
The six rational points we had found above now are
$P_1 = (0,0)$, $-P_1 = (0,-u)$,
$P_2 = (u^2,0)$, $-P_2 = (u^2,-u)$,
$P_3 = (2u^2,0)$, $-P_3 = (2u^2,-u)$
Observe that $P_1 + P_2 + P_3 = 0$; the points $P_1$ and $P_2$
generate a subgroup of $E({\mathbb Q}(u))$ with rank $2$.

A simple calculation shows
$$ 2(0,0) = (4u^6 + 3u^2, -8u^9 - 6u^5 - u), $$
$$ 2(0,-u) = (4u^6 + 3u^2, 8u^9 + 6u^5), $$
$$ 2(u^2,0) = (u^6 + u^2, u^9 - u), $$
$$ 2(u^2,-u) = (u^6 + u^2, -u^9), $$
$$ 2(2u^2,0) = (4u^6-u^2,-8u^9 + 6u^5 - u) $$
$$ 2(2u^2,-u) = (4u^6-u^2,8u^9-6u^5) $$
These points provide us with the following parametrized families of integral
points on our surface:
$$(X,Y,Z,W) = (u^5 + u, 2u^4 - 1, u^5 + 2u^3 - u, u^5 - 2u^3 - u) $$
$$(X,Y,Z,W) = (16u^5 + 16u^3 + 5u, 16u^4+12u^2+1, 16u^5 + 24u^3 + 7u, 16u^5 + 8u^3 - u)$$
$$(X,Y,Z,W) = (16u^5 - 16u^3 + 5u, 16u^4-12u^2+1, 16u^5-8u^3-u, 16u^5-24u^3+7u) $$
None of these give us solutions to our original equation, however.

**3. The Fibonacci connection. **
Since $E$ has rank $2$ over the function field ${\mathbb Q}(u)$, the elliptic
curves $E_u$ will have rank $\ge 2$ except for at most finitely many exceptions
(the only one I noticed is $u = 1$). For some families of specializations, the
rank may be higher. This is the case if we take $u = F_n$, the $n$-th Fibonacci
number. In this family, we have a point independent from the $P_j$ listed above,
which means they have at least rank $3$ (except for the finitely many exceptions
mentioned before). The points (modulo typos) are
$$ Q = (F_{2n-2} F_{2n}, F_{2n-2} F_{2n}F_{2n+1}) $$
if $u = F_{2n}$, and
$$ Q = (F_{2n-1}F_{2n+1}, F_{2n} F_{2n+1}^2) $$
if $u = F_{2n+1}$. While this does not seem to help us, I thought I'd mention it anyway since no one expects the Spanish inqui^H Fibonacci numbers in this problem.