Neater solution. (not obvious where I got formula from)
Note that: $(b^2-a^2)^2+(2ab)^2+(2cd)^2-(c^2+d^2)^2=(a^2+b^2+c^2-d^2)(a^2+b^2-c^2+d^2)$.
So if you have a solution $(a,b,c,d)$ to $a^2+b^2+c^2-d^2=0$ then another solution is:
Applying this to your solution $(1,2,2,3)$ gives $(3,4,12,13)$.
Note the order of terms could be juggled around for further applications of this. The only requirement is that $b>a$.
$(3,4,12,13)$ becomes $(24,12,313)$.
$(3,12,4,13)$ becomes $(35,72,104,185)$.
$(4,12,3,13)$ becomes $(128,96,78,178)$.
Ugly Solution. (more obvious where formula comes from)
Pythagoras's theorem is a related problem only involving three terms: $a^2+b^2=c^2$. It has a fundamental solution of $x^2-y^2, 2xy, x^2+y^2$. We can built upon this.
If we tackle your problem $a^2+b^2+c^2=d^2$ then we can start with $a^2+b^2=(x^2-y^2)^2, c=2xy, d=x^2+y^2$. Note that solving $a^2+b^2=(x^2-y^2)^2$ is another case of Pythagoras: $a=k^2-l^2, b=2kl, x^2-y^2=k^2+l^2$. Note that $x^2-y^2=k^2+l^2$ is the same as your original equation to which you have a solution so we can build another from there: $x=3,k=2,l=1,y=2$ (Note we need $k>l$).
So with $k=2,l=1,y=2,x=3$ then we get $a=3,b=4,c=12,d=13$ giving $3^2+4^2+12^2=13^2$.
So in general if you have a solution $(a,b,c,d)$ (with $a<b$) you can form a new solution: $(b^2-a^2,2ab,2cd,c^2+d^2)$.