The righthand side is even, as is $2j^2$, so $i$ must be even. Say $i=2^nm$, where $m$ is odd and $n\ge 1$, and $j=2^k\ell$, where $\ell$ is odd and $k\ge 0$; then $3m^22^{2n}+\ell^2 2^{2k+1}=77\cdot6^{2012}$. Let $a=\min\{2n,2k+1,2012\}$; then $3m^22^{2n-a}+\ell^2 2^{2k+1-a}=77\cdot3^{2012}2^{2012-a}$, where the exponents on the $2$’s are all non-negative, and at least one is $0$. Clearly it’s not possible for exactly two of these exponents to be positive, and since the first two have opposite parity, it’s not possible for all three to be zero, so exactly one is positive and two are zero. The first and third have the same parity, so we must have $2n-a=2012-a=0$ and $2k+1-a>0$. In particular, $a=2012$, and $n=1006$.
To recapitulate, at this point we know that $i=2^{1006}m$ for some odd $m$, and $j=2^k\ell$ for some odd $\ell$ and $k\ge 1006$, and we can let k\,'=k-1006 and divide through by $2^{2012}$ to get 3m^2+2^{2k\,'+1}\ell^2=77\cdot3^{2012}\;.\tag{1}
Now perform a similar analysis considering factors of $3$ instead of $2$. If $3^b$ and $3^c$ are the highest powers of $3$ dividing $m$ and $\ell$, respectively, then $3m^2$ has $2b+1$ factors of $3$, 2^{2k\,'+1}\ell^2 has $2c$, and the righthand side has $2012$. Reasoning as before, we see that this is possible only if $2c=2012$ and $2b+1>2012$, so that $c=1006$ and $b\ge 1006$. Thus, if we set b\,'=b-1006 and divide through by $3^{2012}$, we can further reduce $(1)$ to
3^{2b\,'+1}u^2+2^{2k\,'+1}v^2=77\tag{2}\;,
where $m=3^{2b+1}u$, $\ell=3^{1006}v$, and $u$ and $v$ are not divisible by $2$ or $3$.
Plainly 0\le b\,'\le 1 and 0\le k\,'\le 2, so there are at most six solutions. Moreover, the only possible values for $u$ and $v$ are $1$ and $5$: $7$ is already too large.
If $u=v=1$, the lefthand side is either too large or at most $3^3+2^5=59$, so we get no solutions.
If $u=5$, b\,' must be $0$, and we get the solution $3\cdot25+2\cdot1=77$.
If $v=5$, k\,' must be $0$, and we get the solution $3^3\cdot1+2\cdot25=77$.
The original equation therefore has two solutions in positive integers and therefore eight in integers (since we can change the sign of $i$ or $j$ independently in each solution).
This can probably be done more elegantly; I wrote it up as I was working it out.