How to solve for an Z-Score of a T-Distribution?

David's method is the route I would go if I didn't want to use Quantile for some reason.

Quantile[StudentTDistribution[49], .95]

==> 1.67655

If you really want to use PDF's to demonstrate this you can still use FindRoot by first setting the integration up so that it only accepts numeric input.

f[y_?NumericQ] := NIntegrate[PDF[StudentTDistribution[49], x], {x, -Infinity, y}]

FindRoot[f[y] == 0.95, {y, 1}]

==> {y -> 1.67655}

Have you tried FindRoot? It's a numerical function looking for a root given some starting location. In addition to that, you can get around integrating by using CDF instead of PDF in the first place:

FindRoot[CDF[StudentTDistribution[49], y] == 0.95, {y, 1}]

{y -> 1.67655}

If you're interested, the reason why your first approach doesn't work: NIntegrate cannot use placeholders, it has to evaluate to a number in all cases. What you're trying to do is equivalent to NIntegrate[x*y, {x,0,1}], and the program complains that it does not know the full integrand since y is undefined, therefore it cannot be evaluated. The fact that you've wrapped a Solve, which inserts the missing variable (x in your case) after NIntegrate has been evaluated, does not have any impact on that.